A  COURSE 


THE  ELEMENTARY  PRINCIPLES 


OF 


CHEMISTRY 


FOR  SECONDARY  SCHOOLS 


FOURTH    EDITION 


BY 

B.  W.  McFAKLAND,  C.E.,  PH.D. 

instructor  in  Chemistry 

IN    THE 

NEW   HAVEN   HIGH   SCHOOL 


NEW  HAVEN 

THE  TUTTLE,  MOREHOUSE  &  TAYLOR  COMPANY 
IQIO 


Copyright,   1907 

by 
B.  W.  McFarland. 


A 

THIRTY-FIVE  WEEKS'   COURSE 

IN 

THE    ELEMENTARY    PRINCIPLES    OF    CHEMISTRY 

FOR  SECONDARY  SCHOOLS. 

TO    BE    USED    IN    CONNECTION    WITH    ANY    GOOD    DESCRIPTIVE    CHEMISTRY. 

BY 

B.    W.    McFARLAND,    C.E.,    Ph.D. 
Instructor  in  Chemistry 

IN  THE 

NEW  HAVEN   HIGH   SCHOOL, 
NEW  HAVEN,  CONN. 


239035 


CONTENTS. 


PART  I.    LABORATORY  WORK. 

Page. 

INTRODUCTION. 

Personal  Equipment   I 

Individual  Laboratory  Equipment i 

Equipment  for  general  Use 2 

The  Note  Book 4 

General   Outline   for  Recitation 5 

Specimen   Exercise    6 

EXPERIMENTAL  WORK. 

Abbreviations  used  in  Directions 8 

Construction  and  Operation  of  the  Bunsen  Burner 9 

Working  with  Glass 9 

The  Collection  of  Gases  over  Water  and  some  Properties  of 

Oxygen    10 

Some  Properties  of  Metallic  Oxides n 

Preparation  and  Properties  of  Oxygen  and  Oxides  of  Non- 
metals    12 

Properties  and  Preparation  of  Hydrogen   13 

Electrolysis    16 

The  Reduction  of  Metallic  Oxides  and  the  Preparation  and 

Purification   of   Water    18 

Solution  and  Crystallization 21 

The  Action  of  Acids  on  Metals  and  on  Basic  Oxides 23 

The  Action  of  Acids  on  Bases  and  on  Carbonates 25 

The  Formation  of  Salts  by  the  Mixing  of  two  Solutions  that 
combine  to  form  an  insoluble  Compound;    also  by  fusing 

together  Basic  with  Acidic  Oxides 27 

The  Preparation  and  Properties  of  Chlorine 28 

Preparation  and  Properties  of  Hydrochloric  Acid 29 

The  Preparation  and  Properties  of  Bromine  and  Iodine 31 

Sulphur    34 


viii  Contents. 

Page. 

Passage  of  the  Electric  Current  through  Liquids   139 

Faraday's  Law   141 

A  simple  Concentration  Cell  141 

Thermochemistry    142 

The  Law  of  Dulong  and  Petit 142 

The  Heat  of  Formation  and  Heat  of  Decomposition 142 

Chemical  Equilibrium  and  the  Law  of  Mass  Action  144 

The  Formation  of  Precipitates    148 

The  Addition  of  highly  dissociated  Acids  or  Salts  to  Solu- 
tions of  weak  Acids  150 

Necessity  for  an  Excess  of  Reagent  to  Produce  complete 

Precipitation    151 

Prevention  of  the  Formation  of  Precipitates  151 

Hydrolysis   153 

PART  IV.    MISCELLANY. 

Review  Questions  on  Part  II 155 

Review  Questions  on  Part  III 158 

College  Examination  Questions   160 

Simple  Problems  in  Elementary  Chemistry  168 

General  Review  Exercise  172 

Table  I,  Review  Transformations -. 174 

Table  II,  Natural  Elements  and  Compounds  in  Groups  175 

Review  Exercise 176 

Table  III,  Common  Elements  and  Compounds  177 

Table  IV,  Showing  degree  of  Solubility  of  some  Compounds  178 

Solubility    Rules    179 

Table  V,  Periodic  Arrangement  of  the  Elements  180 

Table  VI,   Complete   List   of   the   Elements,   giving  Atomic- 
Weights,     Molecular    Weights,     State,     Melting     Points, 

Boiling  Points,  Specific  Heats  and  Specific  Gravities 181 

Table  VII,  Heats  of  Formation  of  some  common  Compounds  183 
Table  VIII,  Tables  of  Length,  Weight  and  Volume  184 

LABORATORY  EQUIPMENT. 

Student's  Individual  Equipment  with  approximate  Cost 185 

Equipment  for  general  Use  186 

Additional  Apparatus  187 

Reagents  for  a  Class  of  20  187 

Suggestions  for  the  Laboratory   192 


PREFACE. 


In  publishing  this  book  the  author  is  primarily  influenced 
by  a  desire  to  place  in  the  hands  of  his  own  students  a  set  of 
directions  for  their  laboratory  work,  combined  with  a  concise 
statement  of  those  definitions,  principles,  and  explanations 
which  he  considers  essential  to  an  intelligent  understanding 
of  the  subject. 

The  course  represents  what  is  being  done  in  this  laboratory 
at  present  and  what  has  been  done  for  several  years.  It  is 
in  reality  the  outcome  of  a  good  many  years  of  experience  in 
teaching  the  subject  from  a  variety  of  text  books  without  any 
printed  laboratory  guide  whatever,  combined  with  the  necessity 
of  obtaining  practical  results  in  a  very  limited  time.  The  avail- 
able time  for  the  subject  in  this  school  consists  of  thirty-five 
laboratory  periods  of  eighty  minutes  each  and  seventy  recitation 
periods  of  forty  minutes  each. 

In  developing  the  subject  it  has  been  the  constant  aim  to 
teach  general  principles  and  to  eliminate  as  far  as  possible  the 
burden  resulting  from  the  mere  memorizing  of  a  large  quan- 
tity of  ill-assorted  facts.  With  this  in  view  the  laboratory  work 
is  so  arranged  that  for  the  first  three  months  the  student  meets 
only  such  chemical  changes  as  are  typical  of  the  fundamental 
reactions  of  chemistry.  With  the  exception  of  oxygen,  hydro- 
gen, and  water,  no  substances  are  studied  during  this  time  that 
require  much  descriptive  work  in  recitation.  The  extra  time 
is  spent  in  constant  drill  on  fundamental  definitions,  the  mean- 
ing and  use  of  symbols,  formulas  and  equations,  and  on  the 
subject  of  chemical  equivalence.  As  a  result  of  this  drill  the 
student  becomes  perfectly  familiar  with  the  language  of  chem- 
istry and  has  no  further  serious  trouble  in  remembering,  classi- 
fying and  expressing  chemical  facts. 


x  Preface. 

Beginning  with  the  subject  of  chlorine,  and  continuing 
through  bromine  and  iodine,  the  general  subject  of  oxidation 
is  taken  up,  and  the  use  of  many  oxidizing  agents  is  studied. 

At  each  following  laboratory  exercise  one  or  more  substances 
are  studied.  The  results  of  the  laboratory  work  are  discussed 
and  explained  in  the  next  recitation. 

The  student  is  required  to  take  careful  original  notes  in  a 
suitable  note  book  of  all  results  obtained  in  the  laboratory  and 
to  write  a  composition  in  the  book  on  each  substance  studied, 
according  to  a  specified  outline.  The  material  for  this  he  gets 
partly  from  his  laboratory  work  and  partly  from  his  text  book 
or  other  book  of  reference. 

At  the  second  recitation  following  the  laboratory  work  the 
substances  studied  are  described  according  to  the  outline. 

Exercises  in  performing  examples  and  in  writing  equations 
are  constantly  given  to  a  few  students  during  each  recitation; 
and  whenever  opportunity  offers,  whole  lessons  are  assigned 
from  the  review  work  indicated  in  Tables  I,  II  and  III  in 
Part  IV. 

The  ideal  condition  in  teaching  chemistry  would  be  to  have 
one  teacher  to  each  student.  A  student  under  those  conditions 
would  be  sure  to  like  the  subject  and  do  well  in  it,  for  the 
course  would  be  exactly  adapted  to  his  needs.  To  adapt  the 
teaching  to  a  large  class  is,  however,  the  actual  condition  which 
confronts  the  instructor.  He  must  neither  overwork  the  slow 
student,  nor  allow  the  bright  one  to  be  idle  and  so  to  feel  that 
he  is  getting  little  from  his  subject.  He  must  also  see  that  the 
average  student,  who  is  very  willing  to  work  but  unwilling  to 
search  long  for  the  ideas  needed  in  his  work,  shall  find  those 
ideas  at  hand.  These  problems,  it  is  hoped,  will  be  solved,  to 
a  certain  extent,  by  giving  to  the  student,  both  in  his  laboratory 
work  and  in  his  outside  study,  a  book  which  will  partially  take 
the  place  of  an  instructor  by  his  side. 

On  account  of  the  short  time  available,  the  quantity  of 
material  used  in  the  experiments  is  extremely  small.  In  order 
to  specify  these  quantities,  when  referring  to  dry  reagents,  a 
particular  small  double  horn  spoon,  more  fully  described  in 
Part  IV,  has  been  adopted  in  this  laboratory  and  is  referred 
to  as  small  spoonfuls  "s.s.,"  or  large  spoonfuls,  "l.s." 

All  reference  numbers  refer  to  paragraphs. 


Preface.  xi 

The  use  of  a  pipe  system  for  the  distribution  of  gases  to 
the  students'  tables  need  in  no  way  interfere  with  the  use  of 
the  book  in  laboratories  not  so  equipped,  for  the  experi- 
ments in  which  it  is  used  are  rather  few  and  could  be  readily 
omitted  without  damaging  the  course;  or  other  experiments 
could  be  substituted  for  them. 

No  apparatus  is  used  that  need  take  more  than  one  minute 
to  put  in  place.  All  experiments  that  might  result  in  accidents 
destructive  to  apparatus  or  material  have  been  avoided,  as  well 
as  those  that  frequently  give  unsatisfactory  results  in  the  hands 
of  students. 

The  course  assumes  that  the  student  has  had  a  good  high 
school  course  in  physics,  and  is  intended  to  be  used  in  the 
fourth  or  senior  year.  It  is,  however,  used  to  some  extent  by 
second  and  third  year  classes  in  this  school,  and  the  results 
are  satisfactory. 

In  conclusion  the  author  wishes  to  express  his  thanks  to 
Dr.  P.  T.  Walden  and  Dr.  H.  W.  Foote,  both  of  the  Sheffield 
Scientific  School,  for  many  valuable  suggestions  and  kindly 
criticisms. 

It  would  give  the  author  pleasure  to  correspond  with  teachers 
who  may  use  the  book,  and  to  receive  from  them,  or  others 
interested,  suggestions  or  criticisms. 


TO  THE  TEACHER. 


The  following  course  may  be  considered  to  be  divided  into 
three  distinct  parts.  The  first  part  has  to  do  entirely  with 
teaching  the  language  of  chemistry,  i.  e.  the  meanings  and  uses 
of  symbols,  formulas  and  equations  as  well  as  the  fundamental 
reactions,  and  lasts  about  twelve  or  fourteen  weeks.  In  addi- 
tion to  the  first  twelve  laboratory  exercises  the  student  is 
thoroughly  drilled  on  everything  in  Part  II  to  art.  65. 
Especial  stress  should  be  laid  upon  the  following  topics: 

1.  Symbols  and  names  of  the  common  elements  in  groups. 

2.  Full  meaning  of  a  formula  of  a  compound. 

3.  Combining  powers  or  valence  of  the  different  elements. 

4.  Formulas  of  binary  compounds. 

5.  How  many  grams  of  one  element  would  be  chemically 
equivalent  to  a  certain  number  of  grams  of  another  element? 

6.  Full  meaning  of  a  chemical  equation. 

7.  Equations   showing  the   formation  of  binary  compounds 
from  the  elements. 

8.  Symbols  and  names  of  the  radicals. 

9.  Formulas  and  names  of  the  compounds  of  all  the  positive 
radicals  with  all  the  negative  radicals. 

10.  Seven  general  methods  for  the  preparation  of  salts. 

This  can  be  accomplished  to  the  entire  satisfaction  of  both 
pupil  and  teacher  in  about  twenty-two  recitations. 

As  a  result  of  the  above  drill  the  student  becomes  so  familiar 
with  the  meaning  and  use  of  formulas  and  equations  that  during 
the  rest  of  the  course  he  experiences  no  trouble  whatever  in 
recording,  remembering  and  explaining  the  chemical  changes 
that  come  under  his  observation.  It  also  leads  to  an  almost 
unconscious  knowledge  of  the  general  subject  of  chemical 
equivalence  which  is  of  the  utmost  value. 


To  the  Teacher.  xiii 

The  second  part  of  the  course  lasts  about  twelve  weeks. 
During  this  time  the  non-metals  are  especially  studied  both 
in  the  laboratory  and  recitation  room.  At  the  first  recitation 
following  a  laboratory  exercise  the  experiments  are  thoroughly 
discussed  and  explained,  the  student  being  expected  to  state 
what  he  did  and  what  he  observed  and  to  explain  what  hap- 
pened, as  far  as  possible,  representing  all  changes  by  equations. 

This  recitation  is  as  much  an  instruction  exercise  as  it  is  a 
recitation,  but  the  student  receives  full  credit  for  having  made 
the  observations  that  might  reasonably  be  expected  from  the 
experimental  work  done. 

During  this  portion  of  the  course  the  subject  of  oxidation 
in  the  wet  way  is  thoroughly  taken  up  and  instruction  is  given 
in  forming  equations  representing  oxidation  reactions. 

At  the  second  recitation  following  a  laboratory  exercise,  the 
subject  of  the  laboratory  work,  when  it  is  a  substance,  is  recited 
according  to  a  specified  outline  which  the  student  fills  out  partly 
from  the  laboratory  work  and  partly  from  the  text  book.  This 
results  in  much  repetition  of  certain  ideas  of  great  importance 
which  are  otherwise  apt  to  be  neglected.  In  this  way  the  text 
book  becomes  a  reference  book  and  is  used  intelligently,  the 
burden  of  remembering  mere  print  being  entirely  eliminated. 

Following  this,  the  student  is  required  to  write  an  article  in 
the  form  of  a  report  on  the  substance  studied,  covering  the  main 
properties  and  describing  and  explaining  what  was  done  with 
it  in  the  laboratory. 

This  provides  an  immediate  review  of  what  was  learned  in 
the  two  recitations  and  requires  an  additional  reading  of  the 
matter  in  the  text  book;  the  whole  constituting  a  very  desir- 
able preparation  for  the  next  laboratory  exercise  at  which  the 
composition  is  due. 

In  the  last  part  of  the  course  the  laboratory  study  of  the 
metals  is  completed,  and  during  the  recitations  the  special 
review  exercises  indicated  in  Tables  I,  II  and  III  are  taken  up. 
The  exercise  indicated  in  Tables  II  and  III  is  of  the  greatest 
value  in  giving  a  review  of  the  whole  subject.  In  this  exercise 
the  student  is  expected  to  represent  by  equations  the  transfor- 
mation of  natural  elements  and  compounds  into  common  com- 
pounds and  elements. 


xiv  To  the  Teacher. 

In  this  work  he  is  not  allowed  to  make  use  of  the  formula 
of  any  substance  whose  formation  he  has  not  already  shown 
from  the  natural  substances.  This,  again,  is  an  exercise  that 
gives  great  repetition  of  the  very  points  most  needed  and  brings 
out  strongly  the  relative  importance  of  many  processes  and 
principles. 

During  the  last  two  or  three  weeks  all  recitations  are  stopped 
and  the  student  is  required  to  spend  all  the  time  in  the  labora- 
tory working  on  the  identification  of  common  chemical  com- 
pounds. In  this  work  he  shows  a  remarkable  amount  of 
interest  for  the  end  of  the  year. 

As  a  review  of  the  laboratory  work  of  the  year  the  exercise 
is  unexcelled. 

Examples  should  be  constantly  given.  At  first  only  those 
involving  weighty  then  those  involving  volumes  of  gases,  finally 
those  involving  volumes,  specific  gravities  and  percentage 
content  of  liquids. 


PART  I. 
LABORATORY  WORK. 


INTRODUCTION. 

Personal  Equipment. 

Each  student  should  provide  himself  with  an  apron,  a  towel, 
a  piece  of  cloth,  a  sponge,  and  a  piece  of  soap. 

The  apron  may  be  best  made  of  denim,  perfectly  plain 
with  a  fixed  strap  around  the  neck  and  a  button  strap  around 
the  waist.  It  should  be  long  enough  to  reach  within  four  inches 
of  the  floor. 

Individual  Laboratory  Equipment. 

Learn  the  names  of  these  things  the  first  day  in  the  laboratory 
and  the  places  where  they  belong. 

To  be  kept  in  drazucr. 

%  lb.  glass  tubing1, 
i  funnel  tube, 
i  combustion  spoon, 
i  glass  bend,  45  cm.  long. 
.  3  pieces  of  copper  wire. 

f  i  piece  of  rubber  tubing. 

i  file. 

i  pair  of  forceps. 

i  glass  stirring  rod. 
t  i  iron  wire  loop. 

First  \  8  squares  of  glass, 

square.         )  3  rubber  connectors. 


Long 
compartment. 


Short 
compartment. 


Second 
square. 


f  i  porcelain  mortar  with  pestle. 
^  3  No.  8,  2-hole  rubber  stoppers, 
[  i  No.  3,  i -hole  rubber  stopper. 


Large 
section. 


3  short  right  angle  bends,  8  cm.  x  8  cm. 

3  medium  right  angle  bends,  8  cm.  x  15  cm. 

i  3"  Royal  Berlin  porcelain  evaporating  dish. 


To  be  kept  in  locker. 

f  i  No.  2  beaker. 
On  the  shelf.   \  i  8  oz.  flask. 

1  funnel. 

f  6  5  oz.  bottles. 

2  500  c.c.  bottles. 
On  the  floor.  <   i  test  tube  rack. 

12  test  tubes  in  rack,  upside  down, 
i  test  tube  holder  on  rack. 


To  be  kept 
in  locker. 


To  be  kept 
in  drawer. 


To  be  kept 
on  table. 


Equipment  for  General  Use. 

'  i  water  pan. 
i  glass  shelf  for  water  pan. 
i  iron  lamp  stand,  two  rings  and  one  clamp. 
i  filter  flask  with  rubber  tubing, 
i  porcelain  sieve  funnel  with  rubber  stopper, 
i  1000  c.c.  bottle. 

i  centimeter  scale. 

i  bulb  test  tube. 

i  package  of  filter  paper. 

i  bundle  of  splints. 

Disks  of  filter  paper  for  porcelain  funnel, 
i  wire  gauze  with  asbestos  center, 
i  iron  crucible 
i  iron  plate, 
i  piece  of  cobalt  glass, 
i  thermometer. 
[  i  package  of  common  wrapping  paper. 

f  i  50  c.c.  graduate,  or  measuring  glass. 
8  reagent  bottles, 
i  incandescent  lamp. 

1  litmus  paper  bottle. 

2  Bunsen  burners  with  tubing, 
i  test  tube  cleaner. 


—3— 

STUDENTS  ARE  EXPECTED  TO  KEEP  THEIR 
SUPPLY  OF  APPARATUS  COMPLETE.  THIS  CAN 
BE  DONE  BY  PAYING  AT  ONCE  FOR  THOSE 
THINGS  THAT  ARE  BROKEN  OR  LOST  THROUGH 
CARELESSNESS,  AND  RECEIVING  NEW  MATE- 
RIAL. 

Care  of  the  Table  and  Lockers. 

SPONGE  OFF  THE  TABLE  AFTER  EVERY  LABO- 
RATORY EXERCISE  WITH  A  MOIST  SPONGE, 
WHETHER  ANYTHING  HAS.  BEEN  SPILLED  ON 
IT  OR  NOT,  AND  WIPE  IT  DRY  WITH  A  CLOTH. 

Keep  the  reagent  bottles  in  the  proper  order,  viz.,  from  left 
to  right — sulphuric  acid,  cone.,  dil. ;  nitric  acid,  cone.,  dil. ; 
hydrochloric  acid,  cone.,  dil. ;  ammonium  hydroxide  and  sodium 
hydroxide. 

Wash  all  the  reagent  bottles  and  sponge  off  the  shelves  once 
a  week. 

Keep  all  of  your  own  apparatus  clean  and  in  the  right  place. 

DON'T  put  the  general  apparatus  in  your  private  locker. 

SOME   LABORATORY    DON'TS. 

Don't  throw  solids  in  the  sink. 

Don't  boil  sulphuric  acid. 

Don't  pour  water  into  hot  sulphuric  acid. 

Don't  aim  a  test  tube  at  your  neighbor.  Point  it  to  the  right, 
down  the  middle  of  the  table. 

Don't  put  anything  into  the  reagent  bottles,  not  even  the 
reagent  itself. 

Don't  mix  chemicals  on  your  own  account.  Students  who 
are  found  doing  this  will  be  dismissed  from  the  laboratory. 

Don't  use  large  quantities  where  small  quantities  will  do  just 
as  well. 

Don't  depend  entirely  on  your  own  observation  for  your  facts ; 
verify  them  by  reading  up  the  subject  in  the  various  reference 
books. 

Don't  pay  any  attention  to  the  way  your  neighbor  is  doing 
his  work. 


—4— 

REMEMBER. 

Neatness  is  of  the  first  importance. 

You  will  be  marked  on  the  way  you  keep  your  desk  and 
locker. 

To  follow  directions  implicitly,  but  try  to  understand  the 
reasons  for  them. 

Where  the  reference  books  are  kept. 

You  are  responsible  for  the  results  whether  you  do  the  work 
or  not. 

To  bring  your  note  book  and  text  book  to  all  exercises, 
both  recitations  and  laboratory  work. 

You  are  at  liberty  to  use  the  laboratory  at  all  times,  when  it 
is  not  occupied  by  a  regular  class. 

To  keep  your  eyes  open. 

To  read  the  signs. 

To  learn  the  names  of  the  common  minerals  and  chemicals 
in  the  wall  cases. 

THE   NOTE   BOOK. 

All  writing  in  the  note  book  is  to  be  in  lead  pencil.  A  hard 
pencil  should  be  kept  for  this  purpose. 

Every  laboratory  exercise  should  appear  in  the  book  in  three 
parts,  as  follows: 

Part  I.  Is  to  contain  whatever  the  instructor  may  have  to 
dictate  before  the  class  goes  to  the  tables.  It  will  contain 
definitions  and  principles  that  are  needed  in  the  day's  work. 

Part  II.  Is  the  most  important  part  of  the  book.  It  is  to 
contain  the  original  notes  of  the  student  on  the  experiment  in 
question. 

The  notes  must  be  short,  on  separate  lines,  contain  a  brief 
reference  to  what  was  done  to  produce  the  result,  and  must 
be  numbered  or  lettered  to  correspond  with  the  directions. 

Part  III.  Is  to  be  a  composition  on  the  day's  laboratory 
work.  The  particular  subject  will  be  assigned  by  the  instructor 
each  day. 

If  the  subject  of  the  composition  is  a  substance,  the  outline 
for  it  will  be  as  follows: 


—5— 


GENERAL  OUTLINE  TO  BE  FOLLOWED  IN  RECI- 
TATION AND  IN  WRITING  COMPOSITIONS. 


Appearance  and  state. 


Occurrence  in  the  free  condition. 


Where  ? 

In  what  quantity  ? 


Occurrence  in  the  combined  condition. 


{  Where  ? 
"I  As  what  ? 


Physical 
Properties. 


Chemical 
Properties. 


Preparation. 

I 

Common  com-  f 
pounds  not 

already 
described. 

Uses.          -! 


Can  it  be  liquefied,  solidified,  or  gasified,  and 

under  what  conditions  ? 
Will  it  dissolve  in  water  or  in  other  liquids  ? 

To  what  extent  ? 

At  what  temperature  does  it  melt  or  boil  ? 
How  does  its  volume  change  with  a  change 

of  temperature  ?     101 

How  does  its  volume  vary  with  the  pressure  ? 
What  is  its  specific  gravity  ? 
If  it  is  a  gas,  what  does  a  liter  weigh  ? 

Will  it  burn  or  support  combustion  ? 

Can  it  be  decomposed  ? 

Does  it  combine  with  oxygen  ? 

What  oxides  does  it  form  ? 

Are  these  oxides  acidic  or  basic  ? 

Does  it  form  any  compounds  with  hydrogen? 

What  experiments  might  be  done  with  it  ? 

Describe  them  in  detail. 

Give  all  equations  involved. 

The  common  way  it  is  made  in  the  laboratory. 
The  common  way  it  is  manufactured. 
Two  other  ways. 
Give  all  equations. 

The  chemical  name  and  formula. 

The  common  name  and  the  mineral  name. 


In  manufacturing. 
In  medicine. 


The  common  test  for  the  substance,  if  it  is  an  element. 


-6— 

NOTE.  It  is  not  expected  that  answers  will  be  found  for  every 
one  of  these  questions  for  any  one  substance.  Whatever  facts 
are  learned  either  from  the  laboratory  work  or  from  the  text 
book  are  to  be  fitted  into  the  outline. 


SPECIMEN    EXERCISE. 
July  24.     Laboratory  Exercise  No.  49. 

EXPERIMENTS  WITH  MANGANESE. 
Part  I. 

1.  Heat  a  small  quantity  of  manganese  dioxide  in  a  sealed 
tube.     What  gas  is  given  off,  and  what  is  left  in  the  tube?     71. 

2.  Heat  }4   g-   °f  manganese  dioxide  in  a  t.t.  with   5   c.c. 
cone.  HC1.     What  gas  is  given  off?    What  kind  of  a  reaction 
was  this?    75. 

3.  Shake  up  some  manganese  dioxide  with  some  strong  sul- 
phurous acid  water. 

4.  Add  a  small  piece  of  sodium  nitrite  to  a  tube  containing 
54   g-  of  manganese  dioxide  and  5   c.c.  of  dilute  nitric  acid. 
Heat  to  boiling,  then  add  J/£  g.  of  red  lead.     78.     What  com- 
pound of  manganese  is  formed?     Write  all  reactions. 

5.  Fuse  i  g.  of  manganese  dioxide  with  I  g.  of  potassium 
carbonate  and  a  little  potassium  nitrate  in  iron  crucible.     Dis- 
solve out  the  product  with  boiling  water,  filter  and  pass  carbon 
dioxide  into  the  solution. 

Part  II. 

1.  a.  Oxygen  given  off,  shown  by  spark, 
b.  Red  powder  left  in  tube,  Mn3O4. 

2.  Chlorine   given   off   and   manganous   chloride    formed   in 
solution.     This  is  an  oxidation  reaction. 

3.  The  sulphurous  acid  is  oxidized  to  sulphuric  acid,  and  the 
manganese  dioxide  is  decolorized. 

4.  The  sodium  nitrite  with  the  nitric  acid  gives  off  fumes 
of  nitrogen  trioxide,  red ;    these   form  nitrous  acid  with  the 
water,  and  this  is  oxidized  by  the  manganese  dioxide  to  nitric 
acid,  the  manganese  being  changed  into  the  nitrate..   The  red 


lead  oxidizes  the  manganese  nitrate  to  permanganic  acid, 
HMnO4,  which  is  purple  in  solution. 

5.  When  the  two  substances  are  fused  the  compound  is  green. 
This  contains  potassium  manganate. 

When  CO2  is  passed  into  a  solution  of  this,  it  is  changed 
to  a  purple  solution. 

Part  III. 

Manganese  is  a  metal  resembling  iron.     It  does  not  occur  in  Appearance 

Occurrence 

the  free  condition,  but  is  very  abundant  in  the  form  of  oxides, 
the  most  common  one  being  manganese  dioxide  or  pyrolusite. 
It  is  a  hard,  brittle  metal  having  a  grey  color  and  a  high  lustre  physicaproPerti( 
which  it  soon  loses  in  moist  air.     In  almost  all  of  its  physical 
properties  it  resembles  iron,  with  which  it  is  very  commonly 
associated,  both  in  the  ore  and  in  the  finished  metal.     It  forms  Chemicpar10pertif 
several  oxides,  of  which  the  lowest  are  basic,  forming  salts  with 
acids,  and  the  highest  acidic,  forming  manganic  and  perman- 
ganic acids. 

When  manganese  dioxide  is  heated,  it  gives  up  part  of  its 
oxygen  and  becomes  a  brown  substance  having  the  formula 
Mn3O4.  If  manganese  dioxide  be  heated  with  hydrochloric 
acid,  some  of  its  oxygen  unites  with  the  hydrogen  of  the  acid, 
thereby  setting  free  chlorine.  This  is  the  common  method  of 
preparing  chlorine  in  the  laboratory.  The  dioxide  itself  is 
reduced  to  the  basic  manganous  oxide,  MnO,  and  this  unites 
with  some  of  the  hydrochloric  acid  present  and  forms  mangan- 
ous chloride.  Sulphurous  acid  is  oxidized  to  sulphuric  acid  by 
manganese  dioxide,  the  manganese  going  into  solution,  partly 
as  manganese  dithionate  and  partly  as  manganese  sulphate. 
If  sodium  nitrite  be  treated  with  nitric  acid,  it  will  be  changed 
to  sodium  nitrate,  and  nitrous  acid  will  be  set  free.  If  man- 
ganese dioxide  be  present,  the  nitrous  acid  will  reduce  it  to 
manganous  oxide,  becoming  changed  itself  to  nitric  acid.  The 
manganous  oxide  unites  with  the  nitric  acid  and  becomes 
manganous  nitrate.  If  some  red  lead  be  added  to  this  solu- 
tion, boiling  hot,  the  manganese  will  be  oxidized  to  per- 
manganic acid,  which  colors  the  liquid  purple.  When  man- 
ganese dioxide  is  fused  with  potassium  carbonate,  it  is  changed 
to  potassium  manganate.  This  is  green  in  color.  And  if  this 


be  dissolved  in  water,  filtered,  and  carbon  dioxide  passed  in,  it 
is  changed  to  potassium  permanganate.  Equations  for  these 
reactions  are  as  follows: 

3MnO2  heated  =  Mn3O4  +  O2. 
MnO2  +  4HC1  =  MnCl2  +  C12  +  2H2O. 

MnO2  +  H2SO3  =  MnSO4  +  H2O. 
MnO2  +  2H2SO3  =  MnS2Oe  +  2H2O. 
NaNO2  +  HNO3  =  NaNO3  +  HNO 


2. 


HNO2  +  MnO2  =  MnO  +  HNO8. 


=  Mn(NO8)2+  H2O. 
26HNO3  +  2Mn(NO3)2  +  5Pb3O4  =  2HMnO4  + 
i5Pb(N03)2  +  i2H20. 

Manganese  can  be  prepared  by  reducing  the  oxide.  This  can 
be  done  with  carbon  at  a  high  temperature.  Or  the  dioxide 
may  be  mixed  with  the  proper  quantity  of  aluminum  and  the 
mixture  ignited.  The  aluminum  unites  with  the  oxygen  and 
the  manganese  is  set  free  in  the  liquid  condition.  A  very  high 
temperature  is  produced  by  the  reaction. 

MnO2  +  C  =  CO2  +  Mn. 
3MnO2  +  4A1  =  2A1203  + 


compounds. 

agent. 


The  metal  is  used  to  alloy  with  iron  and  steel. 
Manganese  dioxide  or  black  oxide  of  manganese,  MnO2,  is 
used  in  the  laboratory  to  prepare  chlorine. 

Potassium  permanganate  is  a  common  laboratory  oxidizing 


The  element  is  most  easily  recognized  by  melting  a  little  of 
the  material  to  be  tested,  with  sodium  carbonate  and  potassium 
nitrate,  on  platinum  foil.  If  any  manganese  is  present,  the 
green  color  of  sodium  manganate  will  be  seen. 


LABORATORY    WORK. 

Abbreviations  used  in  the  directions. 

c.c.,  cubic  centimeters, 
g.,  grams. 

s.s.,  a  small  horn  spoonful.     The  spoon  holds  about  il/2  g. 
of  mercuric  oxide. 


The  spoon  holds  about  9  g.  of 


l.s.,  a  larger  horn  spoonful, 
mercuric  oxide, 
t.t.,  test  tube, 
ppt.,  precipitate. 


LABORATORY  EXERCISE  NO.  i. 

CONSTRUCTION  AND  OPERATION  OF  THE  BUNSEN  BURNER. 

i.  Note  the  general  construction  of  it.  2.  Take  it  apart  and 
make  a  drawing  of  each  piece.  3.  Describe  how  to  light  it. 
4.  Four  ways  in  which,  sooner  or  later,  it  will  be  found  burn- 


Side  view. 


Fig.  i. 


Cross  section. 


ing.  Only  one  correct  way.  5.  How  to  correct  it  when  it  is 
found  burning  at  the  base.  6.  Make  a  drawing  of  the  Bunsen 
flame.  Note  two  parts  to  it.  7.  Heat  a  piece  of  copper  wire  in 
and  about  the  flame.  68.  From  this  decide  what  is  the  hottest 
point  in  the  flame.  Is  the  flame  hollow  in  the  middle? 

LABORATORY  EXERCISE  NO.  2. 

WORKING  WITH  GLASS. 

1.  To  cut  glass  tubing,  make  a  short  sharp  scratch  across 
the  tube  with  a  file,  place  the  file  on  the  table  directly  under 
the  scratch,  with  the  scratch  uppermost,  then  bear  down  gently 
on  the  tube  so  as  to  break  it. 

2.  To  round  the  edges  of  a  glass  tube,  place  the  end  of  the 
tube  in  the  hottest  part  of  the  flame  and  rotate  it  slowly.     Hold 
the  other  end  lower  down. 


— 10 

3.  To  seal  the  end  of  a  tube,  hold  the  end  in  the  hottest 
part  of  the  flame,  slowly  rotating  it  until  it  seals  up. 

4.  To  make  a  bulb  tube,  seal  the  end  of  a  tube  as  above,  then 
remove  it  from  the  flame  and  blow  carefully  at  once  until  the 
bulb  is  of  the  required  size. 

5.  Ordinarily,  to  make  sealed  tubes,  take  a  piece  of  tubing 
that  is  just  twice  as  long  as  the  desired  tube  and  seal  it  off  in 
the  middle,  thereby  making  two  sealed  tubes  at  once. 

6.  To  bend  glass  tubing,  heat  it  lengthwise  in  an  ordinary  gas 
flame  until  it  is  quite  soft;   then  remove  it  from  the  flame  and 
bend  it  quickly  to  the  required  curve.     Do  not  attempt  to 
bend  it  after  the  glass  begins  to  harden. 

What  new  observations  have  you  made  during  this  exercise? 


LABORATORY  EXERCISE  NO.  3. 

THE  COLLECTION  OF  GASES  OVER  WATER  AND  SOME  PROPERTIES 

OF  OXYGEN. 

Materials.     Magnesium  ;  oxygen  ;  splints ;  magnesium  oxide. 

Apparatus.     Water  pan ;   bottles ;    delivery  tube. 

Definitions — Acid  55;  Alkali  57;  Basic  oxide  60;  Alka- 
line oxide  62;  Acidic  oxide  61. 

A  delivery  tube  is  any  combination  of  rubber  or  glass  tubing 
by  means  of  which  a  gas  can  be  transferred  from  one  place  to 
another. 

A  water  pan  is  used  for  the  collection  of  gases  over  water. 

1.  Fill   several  bottles   with   oxygen.      Holding  a   splint  by 
means  of  the  forceps,  ignite  it  and  thrust  it  down  into  one  of  the 
bottles.     68. 

2.  Repeat  with  magnesium  ribbon.     68.     Burn  several  pieces. 
Scrape  some  of  the  product  out  on  a  glass  plate  and  find  out. 
whether  it  reacts  acid  or  alkaline  toward  moist  litmus  paper. 
62. 

3.  Test  the  action  of  magnesium  oxide  in  water  and  in  dilute 
acids.     60.     In  which  does  it  dissolve  readily  ?     Is  it  a  basic  or 
an  acidic  oxide? 

4.  Make  very  dilute  solutions  of  each  of  the  reagents  on  the 
shelf,  in  test  tubes.     Test  these  solutions  with  litmus  paper. 
Which  are  acid,  and  which  are  alkaline? 


— II — 


Never  lay  the  stopper  of  a  bottle  down  on  the  table. 
Turn  the  hand  over,  take  the  stopper  between  the  first  and 
second  fingers,  remove  it  from  the  bottle,  pick  up  and  use 
the  bottle  with  the  same  hand,  then  replace  the  stopper. 


LABORATORY  EXERCISE  NO.  4. 

SOME    PROPERTIES    OF    METALLIC    OXIDES. 

Materials.  Sodium  wire,  made  before  the  class;  mercuric 
oxide,  HgO;  splints;  manganese  dioxide,  MnO2;  red  oxide 
of  lead,  Pb3O4 ;  calcium  oxide,  CaO ;  magnesium  oxide,  MgO. 

Apparatus.  Iron  crucible;  lamp  stand;  funnel;  filter 
paper. 

Definitions — Properties  of  matter.     5. 

Filtrate.     The  liquid  which  runs  through  a  filter. 

Residue.     The  material  left  on  the  filter. 

Note.  A  solution  is  always  transparent.  If  a  liquid 
is  not  transparent  it  probably  contains  a  finely  divided  solid 
that  remains  in  suspension  and  could  be  filtered  off. 


How  to  use  the  lamp  stand. 


To  be  demonstrated 


How  to  make  a  filter.  ,   ,      , 

TT  r  by  the 

How  to  filter. 

TT        f     •    •  ,  instructor. 

Use  of  stirring  rod. 

1.  Place  a  small  piece  of  sodium  on  a  piece  of  wet  filter  paper. 
Turn  up  the  edges  of  the  paper  so  the  melted  sodium  cannot 
roll  off.     The  white  smoke  is  sodium  oxide. 

2.  Heat  a  piece  of  sodium  wire  in  iron  crucible  on  lamp  stand 
until  no   further  change  takes  place.     Let  it  cool   for  half  a 
minute,  add  water,  heat  to  boiling  and  filter  into  t.t.     Feel  the 
filtrate  with  the  fingers  and  note  its  action  on  litmus  paper. 
What  is  an  alkaline  oxide?     62. 

3.  Place  a  small  quantity  of  mercuric  oxide  in  a  sealed  tube, 
not  more  than  tf>  of  a  s.s.,  heat  and  test  the  escaping  gas  with 
a  spark  on  the  end  of  a  splint.     What  is  left  in  the  tube?     71. 

4.  Heat  manganese  dioxide  in  sealed  tube,  and  test  escaping 
gas  for  oxygen.     71. 

5.  Repeat,  using  red  lead.     78-7. 


12- 


6.  Test  the  solubility  of  each  of  the  oxides  mentioned  at  the 
beginning  of  this  exercise,  in  the  dilute  acids,  and  tabulate  the 
results.  Which  are  alkaline  oxides?  62.  Which  are  basic? 
Which  one  is  not  a  basic  oxide?  Do  all  oxides  give  oxygen 
when  heated? 

LABORATORY  EXERCISE  NO.  5. 

PREPARATION  AND  PROPERTIES  OF  OXYGEN  AND  OXIDES  OF 
NON-METALS. 

Materials.  Manganese  dioxide,  MnO2 ;  oxygen  mixture 
(l/2  MnO2  and  */2  KC1O3)  ;  charcoal,  C;  sulphur,  S;  phos- 
phorus, P  176;  arsenic  trio'xide,  As,O3. 

Apparatus.  Water  pan ;  lamp  stand ;  bottles ;  combustion 
spoon  ;  iron  plate ;  delivery  tube. 

Definitions — Acidic  oxide  61;   Acid  anhydride  63. 

I.  Fit  a  t.t.  with  a  rubber  stopper  and  a  delivery  tube.  Place 
in  the  tube  about  one  inch  of  the  oxygen  mixture.  Fix  the 


Fig.  2. 

tube  in  the  clamp  of  the  lamp  stand,  in  a  nearly  horizontal 
position  as  in  fig.  2.  The  mouth  of  the  tube  should  be  a  little 
lower  than  the  heated  portion.  Heat  the  mixture,  being 


—13— 

careful  not  to  melt  the  tube,  and  collect  the  gas  in  bottles 
in  the  water  pan.  71.  If  this  does  not  give  enough  gas  to 
fill  all  the  bottles,  fill  the  rest  from  the  supply  pipe. 

2.  Balance  a  small  piece  of  charcoal  on  the  combustion  spoon, 
ignite  it  and  lower  it  into  a  bottle  of  oxygen.     68.     When  it 
will  burn  no  longer  add  a  small  quantity  of  water  to  the  bottle, 
cover  it  with  the  palm  of  the  hand  tightly,  and  shake  the  water 
about  in  the  bottle.     What  kind  of  oxides  do  most  non-metals 
form?     61-63.     Test  this  water  with  litmus. 

3.  Pick  up  a  small  bead  of  phosphorus  with  the  forceps  and 
dry  it  by  touching  it  to  filter  paper.     Place  it  in  the  cold  dry 
combustion   spoon.     Ignite  it  in  the  hood  and  lower  it  in  a 
bottle  of  oxygen  carried  there  for  the  purpose.     What  is  the 
white  smoke?    What  becomes  of  it?    Add  water  to  the  bottle, 
shake  and  test  with  litmus  paper. 

4.  Ignite  some  sulphur  in   combustion  spoon  and  lower   it 
into  a  bottle  of  oxygen.      Add  water  and   shake.      Test  the 
water  with  litmus  paper. 

5.  Place  a   dry  piece  of  phosphorus   on  a  dry   iron   plate; 
ignite  it  and  cover  it  at  once  with  a  large  dry  bottle.      Let 
it  stand  until  the  fumes  have  settled.     These  fumes  are  phos- 
phorus pentoxide.     Scrape  some  of  the  oxide  out  on  a  piece  of 
glass.     Breathe  on  it.     Add  one  drop  of  water  to  it.     Test  the 
product  with  litmus  paper. 

6.  Test  arsenic  trioxide  with  litmus  paper.     See  if  it  will 
dissolve  in  an  alkali.     What  kind  of  an  oxide  is  it? 

LABORATORY  EXERCISE  NO.  6. 

PROPERTIES  AND  PREPARATION  OF  HYDROGEN. 

Materials.     Granulated  zinc ;    splints. 

Apparatus.      Test    tube ;     one-hole    rubber    stopper ;     glass 
bends ;    rubber  connectors ;    water  pan ;    bottles ;    glass  plates. 

1.  Making  use  of  the  water  pan,   fill  all  the  bottles  with 
hydrogen    from   the   supply   pipe.      Keep   them   inverted   and 
covered  with  glass  plates. 

2.  Does  hydrogen  burn?     Attach  a  delivery  tube  to  the  sup- 
ply pipe,  fit  the  end  with  a  glass  tip,  turn  on  a  slow  stream  of 
gas  and   attempt  to   light  it.     What   color  has   the  hydrogen 
flame  ? 


3.  Does  it  support  combustion?     Place  one  of  the  covered 
bottles  upside  down  on  a  ring  of  the  lamp  stand,  which  will 
support  the  bottle  when  the  glass  plate  is  removed.     Fasten 
a  splint  on  a  wire.     Now  remove  the  glass  plate  and  support 
the  bottle  on  the  ring,  covering  it  again  at  once  with  the  glass 
plate  held  in  the  left  hand.     Ignite  the  splint  and  quickly  plunge 
it  up  into  the  bottle  the  instant  the  plate  is  removed.     Withdraw 
it  slowly.     What  do  you  see? 

4.  Does  this  gas  dissolve  in  water?     Add  about  an  inch  of 
water  to  a  bottle  of  hydrogen,  cover  the  bottle  tightly  with  the 
palm  of  the  hand,  and  shake  it  thoroughly.     If  the  gas  is  soluble 
in  water,  there  will  be  a  slight  suction  on  the  hand.      v 

5.  Is  hydrogen  lighter  than  air?     Fill  two  bottles  with  the 
gas.     Remove  the   cover  of   one  and  leave   it   standing  right 
side  up  for  one  minute.     Try  to  ignite  the  gas  in  the  bottle  at 
the  end  of  that  time.     Uncover  another  bottle,  but  keep  it  upside 
down  for  one  minute  (this  can  best  be  done  by  supporting  it 
on  the  lamp  stand).     Then  ignite  the  gas.     Obtain  a  clay  pipe, 
and  blow  soap  bubbles  with  hydrogen  from  the  supply  pipe. 
Have  a  bottle  full  of  hydrogen  upside  down  on  a  bottle  of  the 
same  size  full  of  air.     Invert  the  two  bottles  and  after  three 
minutes  test  them  both  to  see  which   contains  the  hydrogen. 

6.  Is  a  mixture  of  hydrogen  and  air  explosive?     Mix  hydro- 
gen and  air  in  various  proportions  and  attempt  to  ignite  the 
mixture.     If  it  is  explosive,  ascertain  what  the  most  explosive 
mixture  is. 

7.  What  is  formed  when  hydrogen  burns?     Insert  a  small 
jet  of  burning  hydrogen  in  a   dry  bottle.      Is   any  moisture 
noticeable  ? 

8.  Hydrogen  does  not  occur  free.     It  must  be  separated  from 
some  of  its  compounds.     How  can  it  be  obtained  from  water? 
It  can  most  easily  be  obtained  from  acids  in  the  laboratory  by 
putting  some  metal  with  them.     64a.     Pour  some  dilute  sul- 
phuric acid  on  some  granulated  zinc  in  a  test  tube.     How  could 
you  collect  this  gas  ?     Fit  the  tube  with  a  rubber  stopper  and  a 
delivery  tube  and  collect  several  bottles  of  the  gas.     Prove  that 
this  gas  is  the  same  as  that  with  which  you  have  just  been 
experimenting  and  which  is  known  to  be  hydrogen.     In  how 
many  ways  could  you  prove  it  to  be  the  same? 


—15— 

In  case  it  is  desired  to  generate  a  larger  quantity  of  hydrogen 
than  can  well  be  obtained  from  a  test  tube,  the  apparatus  indi- 
cated in  fig.  3  should  be  used.  The  funnel  tube  is  passed 
through  a  ring  of  the  lamp  stand  simply  to  prevent  the  apparatus 
from  being  knocked  over.  It  goes  to  the  bottom  of  the  250 
c.c.  bottle,  A,  containing  one  or  two  inches  of  granulated  zinc. 
A  short  bend  comes  out  of  the  two-holed  rubber  stopper  in 
A  and  is  connected  by  a  rubber  connector  to  a  medium  bend 
that  passes  through  another  rubber  stopper  and  goes  to  the 
bottom  of  the  wash  bottle,  B,  which  is  three  quarters  full 


of  ivater.  Enough  water  is  now  poured  through  the  funnel 
tube  to  cover  the  zinc  in  A.  Cone,  sulphuric  acid  is  then  added 
little  by  little,  very  cautiously,  until  the  desired  reaction  is 
obtained.  Sometimes  the  zinc  is  too  pure  to  react  properly  with 
pure  acid,  in  which  case  the  action  will  start  readily  if  a  little 
copper  sulphate  solution  be  added. 

Never  ignite  a  jet  of  hydrogen  coming  from  a  generator 
in  operation. 

9.  When  the  zinc  comes  in  contact  with  the  sulphuric  acid, 
each  atom  of  zinc  displaces  two  atoms  of  hydrogen  from  the 
acid  and  forms  a  compound  called  zinc  sulphate.  64a.  Cover 


the  bottom  of  the  beaker  with  granulated  zinc.  Cover  the  zinc 
with  dilute  sulphuric  acid.  Put  this  away  in  your  locker  until 
the  next  time.  What  do  you  find  besides  some  undissolved 
zinc  ?  64a. 

10.  Fasten  a  glass  jet-tube  20  cm.  long,  drawn  out  to  a  small 
hole,  vertically  in  the  clamp  of  the  lamp  stand,  connect  this  with 
the  supply  pipe,  turn  on  a  small  stream  of  hydrogen,  and  ignite 
the  jet.  Put  slowly  down  over  this  jet,  glass  tubes  of  various 
sizes.  The  noises  are  called  hydrogen  tones. 

LABORATORY  EXERCISE  NO.  7. 

ELECTROLYSIS. 

Materials.  Sodium  chloride;  potassium  sulphate;  litmus 
solution. 

Apparatus.     Voltameter. 

Definitions — 

Electrolysis,  the  process  of  decomposing  a  dissolved  or 
melted  substance  by  means  of  the  electric  current, 

Electrolyte.  A  substance  which,  when  dissolved  or  melted, 
will  allow  the  passage  of  the  electric  current. 

Electrodes.  The  terminals  immersed  in  a  liquid  by  means 
of  which  the  current  enters  and  leaves. 

Positive  electrode.     The  one  on  which  the  oxygen  appears. 

Negative  electrode.  The  one  toward  which  the  current 
travels  and  on  which  the  hydrogen  appears. 

Questions  to  be  answered  by  experiment. 

1.  Will  distilled  water  allow  the  current  to  pass  through  it? 

2.  Will  hydrant  water  allow  the  current  to  pass  through  it? 

3.  What  effect  has  a  small  quantity  of  sulphuric  acid  on  the 
conductivity  of  distilled  water? 

4.  Will  the  current  pass  through  a  very  dilute  solution  of 
sodium  hydroxide?     124. 

5.  How  does  the  presence  of  salt  of  any  kind  affect  the 
passage  of  the  current?     124. 

6.  What  will  be  the  effect  of  passing  a  current  through  a  very 
dilute    solution    of    potassium    sulphate    colored    with    litmus 
solution?     124. 

7.  What   classes   of   substances,   when   dissolved,   allow   the 
passage  of  the  current  through  water? 


—17— 

1.  Clean  the  voltameter  by  taking  the  glass  part  out  of  the 
base  and  running  hydrant  water  through  it,  then  rinse  with 
distilled  water.     Fill  it  with  distilled  water  and  replace  it  in 
the  support.     Turn  on  the  current,     (no  volt  current  through 
one  32  c.p.  lamp.)     Judge  of  the  amount  of  current  passing 
by  the  rate  at  which  hydrogen  is  generated. 

2.  Repeat  exactly,  using  hydrant  water.     Note  that  in  each 
case  very  little  current  passes  but  that  much  more  gets  through 
the  hydrant  water  than  through  the  distilled  water. 

3.  Make  a  mixture  of  100  c.c.  of  water  and  5  c.c.  of  cone, 
sulphuric  acid.     Pour  this  into  the  apparatus  until  the  level  of 


Voltameter. 
Fig.  4. 

the  liquid  comes  nearly  to  the  top  of  the  side  tubes.  Have  the 
stop-cock  at  A  open  and  fit  solid  rubber  stoppers  into  the  top 
of  the  side  tubes  so  there  will  be  no  air  inclosed.  Then  turn 
on  the  current.  Note  the  time.  Let  the  current  run  until  one 
tube  is  just  filled  with  a  gas.  Note  the  time.  Shut  off  the 
current  and  close  the  stop-cock  at  A.  Remove  one  of  the 
stoppers  and  test  the  gas  at  once  with  a  flame  on  a  splint. 
Replace  the  stopper  and  test  the  gas  on  the  other  side.  Which 
was  hydrogen?  How  much  hydrogen  was  there  compared  to 
oxygen?  Find  what  volume  of  hydrogen  was  formed  in  the 
measured  time.  The  current  you  had  was  about  .8  of  an 
2 


— 1 8— 

ampere.  Find  how  many  c.c.  of  hydrogen  one  ampere  would 
give  in  one  second.  One  ampere  should  give  exactly  .0000103 
g.  of  hydrogen  in  one  second. 

4.  Test  solutions  of  sodium  hydroxide  and  sodium  chloride 
in  the  same  way.     What  effect  have  acids,  bases  and  salts  in 
solution  on  the  conductivity  of  water?     118.     Why  did  hydrant 
water    probably    conduct    the    current    better    than    distilled 
water?     124. 

5.  Make  a  solution  of  pure  potassium  sulphate  and  color  it 
by  the  addition  of  some  litmus  solution.     124.     The  voltameter 
must  be  thoroughly  washed  before  adding  the  liquid.     Turn 
on  the  current  and  note  the  effect.     When  oxygen   salts  of 
sodium  or  potassium  are  electrolyzed  an  acid  is  formed  on  one 
side  and  an  alkali  on  the  other. 


LABORATORY  EXERCISE  NO.  8. 

THE  REDUCTION  OF  METALLIC  OXIDES  AND  THE  PREPARATION 
AND  PURIFICATION  OF  WATER. 

Materials.  Copper  scale,  a  mixture  of  cupric  oxide,  CuO, 
and  cuprous  oxide,  Cu2O ;  lead  oxide  or  litharge,  PbO ;  potas- 
sium permanganate  solution,  KMnO4;  crystallized  copper  sul- 
phate or  blue  vitriol,  CuSO4  +  5H2O ;  copper  gauze. 

Apparatus.  Lamp  stand;  wire  gauze;  flask;  No.  8  two- 
hole  rubber  stopper ;  delivery  tube ;  funnel ;  filter  paper ;  sealed 
tubes. 

Definitions — Reduction.     67.     Reducing  agent.     80. 

A  precipitate  is  a  solid  that  separates  from  solution.  In  the 
laboratory  it  is  commonly  formed  by  mixing  two  transparent 
solutions.  WThen  this  is  done,  if  the  mixture  is  not  perfectly 
transparent,  a  precipitate  has  been  formed. 

Crystalline.  A  substance  is  said  to  be  crystalline  when  it 
possesses  naturally  formed  faces  that  reflect  light. 

Amorphous.     Not  crystalline. 

Water  of  crystallization.  Many  substances  when  crystal- 
lizing from  a  water  solution,  combine  chemically  with  a  definite 
number  of  molecules  of  water.  This  water  is  necessary  for  the 
crystalline  form  and  color  of  the  substance. 


—19— 


Efflorescence.  Many  substances  containing  water  of  crys- 
tallization give  up  part  or  all  of  this  water  to  the  air  when 
exposed,  thereby  losing  their  crystalline  form  and  color,  and 
become  amorphous  powders.  Such  substances  are  said  to 
effloresce. 

Distillation.  The  process  of  boiling  a  substance  and  con- 
densing the  vapor. 

Distillate.  The  liquid  that  comes  out  of  a  condenser,  con- 
densed vapor. 


Fig.  5. 

i.  Hydrogen  has  a  very  great  attraction  for  oxygen.  It  will 
not  only  combine  with  it  directly,  but  it  will  extract  it  from 
other  compounds  and  combine  with  it  to  form  water.  80-81. 
Push  a  small  piece  of  copper  gauze,  by  means  of  the  handle 
of  the  combustion  spoon,  a  little  more  than  half  way  through  a 
glass  tube  about  eighteen  centimeters  long.  Add  to  this  half  an 
inch  of  copper  scale,  then  push  in  another  piece  of  gauze. 
Fasten  the  tube  on  a  ring  of  the  lamp  stand  as  in  fig.  5.  Con- 
nect the  tube  to  the  supply  pipe,  and  turn  on  a  very  small 
quantity  of  hydrogen.  Light  the  gas  at  the  other  end  and  turn 
down  the  gas  until  the  flame  is  only  one-half  inch  high.  Now 
slowly  heat  the  copper  scale  with  the  Bunsen  burner.  Record 
all  results  and  represent  by  equations. 


— 20 

2.  Repeat  exactly,  using  lead  oxide  in  place  of  the  copper 
oxide. 

3.  Water  can  be  purified  from  suspended  and  insoluble  sub- 
stances by  filtration.     Dissolve  some  oxygen  mixture  in  a  test 
tube  half  full  of  water  and  filter  into  another  test  tube.     What 
has  been  removed  from  the  water?     Evaporate  the  filtrate  by 


Fig.  6. 


heating  in  the  evaporating  dish  on  the  wire  gauze.     What  is 
the  residue? 

4.  Water  can  be  purified  from  all  kinds  of  matter  by  distilla- 
tion. Fill  the  flask  half  full  of  water.  Add  a  little  sulphuric 
acid  and  some  potassium  permanganate  solution.  Arrange  this 
on  the  wire  gauze  on  the  lamp  stand  with  long  glass  delivery 
tube  as  in  fig.  6.  Heat  to  boiling  and  collect  the  distillate  in  a 
bottle.  Taste  the  water  that  distills  over. 


— 21 — 

5.  Heat  a  small  quantity  of  copper  sulphate  in  a  sealed  tube. 
After  the  material  is  white,  let  some  of  the  water  run  back 
on  it. 

6.  Dissolve    I    s.s.   of   lead   oxide   in    10   c.c.   boiling   dilute 
hydrochloric  acid.     Pour  the   solution   into   water   in   another 
tube.      This  is  a  precipitate  of  lead  chloride,   soluble  in   hot 
water. 

7.  Repeat,  using  copper  scale.     The  precipitate  is  cuprous 
chloride. 


LABORATORY  EXERCISE  NO.  9. 

SOLUTION  AND  CRYSTALLIZATION. 

Materials.  Sodium  sulphate,  Na2SO4  +  ?H2O ;  copper  sul- 
phate, CuSO4  +  5H2O. 

Apparatus.     Test  tubes;    funnel;   filter  paper;   stirring  rod. 

Definitions — A  cold  saturated  solution  is  one  containing  all 
of  a  given  substance  dissolved  in  it  which  it  can  absorb  at  that 
temperature.  It  can  be  made  in  two  ways.  Either  by  shaking 
an  excess  of  the  powdered  substance  with  the  liquid  until  no 
more  will  dissolve  and  filtering  off  the  excess,  or  by  cooling  a 
hot  solution  and  filtering  off  the  substance  that  is  deposited  as  a 
result  of  the  cooling. 

A  hot  saturated  solution  is  one  containing  all  of  a  given 
substance  which  it  can  dissolve  at  that  temperature.  It  is  best 
made  by  gradually  adding  the  powdered  substance  to  the  boiling 
liquid  until  there  is  a  decided  excess  present. 

A  supersaturated  solution.  When  a  hot  saturated  solution 
is  allowed  to  cool,  some  of  the  material  in  solution  generally 
crystallizes  out.  The  amount  that  separates  is  the  exact  amount 
that  the  hot  liquid  dissolves  more  than  the  cold.  Sometimes  the 
excess  refuses  to  crystallize  when  the  hot  saturated  solution  is 
cooled.  The  solution  is  then  said  to  be  supersaturated. 

A  supersaturated  solution  may  be  caused  to  crystallize  either 
by  dropping  in  a  small  crystal  of  the  same  material,  or  by 
scratching  the  inside  of  the  vessel  with  a  glass  rod. 

Mother  liquor.  The  liquid  that  is  left  after  the  crystals  that 
have  separated  from  it  have  been  removed. 


Decant.  To  pour  off  a  liquid  gently  so  as  not  to  disturb  the 
sediment. 

Use  of  filter  flask.  The  filter  flask  is  an  arrangement  for 
rapidly  filtering  off  crystals  or  solid  residues  by  means  of 
suction. 

Fit  the  rubber  stopper  with  the  porcelain  sieve  funnel  tightly 
into  the  flask,  connect  the  rubber  tube  to  the  suction  pump  and 
turn  on  the  water.  Test  the  apparatus  by  putting  the  palm 
of  the  hand  firmly  down  over  the  funnel;  a  decided  suction 
should  be  felt  almost  at  once.  Moisten  a  small  disc  of  filter 
paper  and  fit  it  carefully  over  the  sieve  plate.  Shake  up  the 
material  to  be  filtered  and  decant  it  quickly  into  the  funnel. 
It  will  be  sucked  dry  almost  at  once.  If  necessary,  and  the 
substance  admits  of  it,  it  may  be  washed  free  from  the  mother 
liquor  by  pouring  successive  small  portions  of  water  on  it.  The 
disc  of  filter  paper  with  the  substance  is  then  removed,  placed 
on  a  larger  piece  of  filter  paper  and  set  aside  to  dry. 

PREPARATION  OF  A  SUPERSATURATED  SOLUTION  OF  SODIUM 
SULPHATE. 

1.  Wash  five  test  tubes  and  the  funnel  thoroughly.     Prepare 
a  filter  and  stand  it  in  one  of  the  test  tubes. 

2.  Fill  a  dry  test  tube  one  third   full  of   sodium  sulphate. 
Heat  a  third  of  a  test  tube  full  of  water  nearly  to  boiling. 
Transfer  the  sodium  sulphate  to  the  hot  water,  cover  with  the 
thumb  or  with  a  rubber  stopper  and  shake  thoroughly  for  about 
three  minutes. 

3.  Filter  small  portions  into  the  several  test  tubes.     Do  not 
let  any  drops  touch  the  side  of  the  tube. 

4.  Cover  each  tube  with  a  clean  wet  glass  plate.     Cool  the 
tubes  by  standing  them  in  water  or  by  letting  a  stream  of  water 
run  over  them,  without  shaking. 

5.  If  no  crystals  formed  during  the  cooling,  you  have  suc- 
ceeded in  getting  a  supersaturated  solution  of  sodium  sulphate. 
Into  one  tube  drop  a  small  crystal  of  the  original  salt.     Scratch 
the  inside  of  a  second  with  a  glass  stirring  rod.     Allow  some 
dust  from  the  floor  to  fall  into  a  third.     Simply  shake  a  fourth 
thoroughly. 


—23— 

6.  The  crystals  that  separate  contain  ten  molecules  of  water 
of  crystallization  and  are  known  as  Glauber's  Salt. 

7.  Filter  off  all  of  this  material,  making  use  of  the  filter  flask. 
The  salt  is  very  soluble,  therefore  you  cannot  wash  it.     Set  it 
aside  to  dry  and  explain  next  time  what  has  happened  to  it. 

8.  Pulverize  some  copper  sulphate  and  add  it  gradually  to 
a  t.t.  half  full  of  boiling  water  until,  after  boiling,  a  little  of 
the   salt   remains   undissolved.      Filter   this   hot    solution   into 
several  clean  tubes.     On  cooling,  the  excess  of  salt  will  crys- 
tallize.    If  it  does  not,  it  can  be  made  to  do  so  by  scratching 
the  inside  of  the  tube  with  a  glass  rod. 


LABORATORY  EXERCISE  NO.   10. 

THE  ACTION  OF  ACIDS  ON  METALS  AND  ON  BASIC  OXIDES. 

64-a-b. 

Materials.  Iron  wire;  magnesium  wire;  zinc  oxide;  mag- 
nesium oxide. 

Apparatus.  Lamp  stand;  test  tubes;  funnel;  filter  paper; 
stirring  rod. 

Definitions — An  acid  55.  A  Base  56.  A  salt  58.  A  basic 
oxide  60.  Water  of  crystallization,  see  lab.  ex.  No.  8. 

i.  Fit  a  bulb-test  tube  with  a  rubber  stopper  and  delivery 
tube.  Bend  up  about  5  g.  of  iron  wire  in  a  bunch  two 
inches  long  so  it  will  go  readily  into  the  tube.  Cover 
this  with  cone,  hydrochloric  acid.  Fasten  the  tube  in  the 
clamp  of  the  lamp  stand  at  an  angle  of  forty-five  degrees 
so  the  bottom  of  the  tube  will  come  near  the  base  of  a  Bunsen 
burner  flame  as  in  fig.  7.  Place  the  flame  near  enough  to 
the  tube  so  the  reaction  will  proceed  readily  without  causing 
the  liquid  to  boil  over.  Ignite  the  gas.  Collect  some  of 
it  over  water,  satisfy  yourself  that  it  is  hydrogen.  All  iron 
contains  sulphur  and  phosphorus  as  impurities  and  in  this 
reaction  these  come  out  as  the  corresponding  hydrogen  com- 
pounds, H2S  and  H3P;  both  of  these  have  very  disagreeable 
odors.  To  get  rid  of  them  stand  a  second  burner  near  the  end 
of  the  delivery  tube  to  keep  the  jet  of  gas  burning. 


—24— 

2.  When  the  action  is  nearly  over,  filter  the  liquid  into  a  test 
tube.     The  filtrate  should  be  green.     Cool  and  crystallize,  if 
possible.     If  it  will  not  crystallize,  boil  it  down  until  it  does 
crystallize    on    cooling.       The    product    is    ferrous    chloride, 
FeCl2  +  4H20. 

3.  To  three  c.c.  of  dilute  sulphuric  acid  in  evap.  dish  add 
bits  of  magnesium  wire  until  no  more  will  dissolve.     What  gas 
was  given  off  ?     Boil  a  little,  cool,  and  drop  in  a  small  crystal  of 
magnesium   sulphate.      The   crystals   contain   seven   molecules 
of  water  of  crystallization  and  are  known  as  Epsom  Salt. 


Fig.  7- 

4.  Make  ferrous  sulphate  just  as  you  made  ferrous  chloride, 
but  in  place  of  cone,  acid  use  dil.  sulphuric  acid  with  an  equal 
bulk  of  water.    This  product  is  known  as  green  vitriol  and  has 
the  formula  FeSO4  +  7H2O. 

5.  Make  magnesium  chloride   from  the  metal  just  as  you 
made  magnesium  sulphate.     The  formula  for  the  crystals  is 
MgCl2  +  6H2O. 

6.  Write  twenty-five  equations  showing  the  formation  of  the 
salts  indicated  in  64-a  by  the  above  method. 

7.  Place  about  10  c.c.  of  dilute  sulphuric  acid  in  beaker  on 
wire  gauze  on  lamp  stand,  add  an  equal  amount  of  water,  heat 
nearly  to  boiling  and  add  zinc  oxide  little  by  little  with  much 


—25— 

stirring  until  no  more  -will  dissolve.  Filter  into  t.t.,  add  one  or 
two  drops  of  acid  to  filtrate,  boil  down  and  crystallize.  Why 
was  no  hydrogen  given  off  in  this  case? 

8.  Repeat,  using  magnesium  oxide  and  sulphuric  acid. 

9.  Write   twenty-five   equations,   showing  the   formation   of 
the  same  twenty-five  salts  by  this  method.     64-b. 


LABORATORY  EXERCISE  NO.   n. 

THE  ACTION  OF  ACIDS  ON  BASES  AND  ON  CARBONATES. 
64-c-d. 

Materials.     Sodium  carbonate. 

Apparatus.  Lamp  stand;  wire  gauze;  evaporating  dish; 
stirring  rod ;  test  tubes. 

Definitions — An  alkali.     57.     A  base.     56. 

Neutral,  reacting  neither  acid  nor  alkaline  toward  litmus. 

Neutralization,  the  process  of  rendering  a  substance  inactive 
toward  litmus  paper. 

PREPARATION  OF  A  SALT  BY  NEUTRALIZING  A  BASE  WITH  AN 

ACID.     64-c. 

1.  Place  about  5   c.c.   of  sodium  hydroxide   solution   from 
reagent  bottle  in  porcelain  dish  on  ring  of  lamp  stand.     174. 
Add  an  equal  amount  of  water.     Now  add  dilute  hydrochloric 
acid  little  by  little  until  a  small  piece  of  litmus  paper  just  turns 
red.     It  is  evident,  now,  that  you  have  too  much  acid  present. 
Take  a  small  quantity  of  the  sodium  hydroxide  solution  in  a 
test  tube  and  dilute  it  with  five  times  its  bulk  of  water.     Add 
this  solution  little  by  little  with  much  stirring  until  the  paper 
just  turns  blue.     You  now  have  too  much  alkali.     Dilute  some 
more  of  the  hydrochloric  acid  with  ten  times  its  bulk  of  water 
and  add  it  very  cautiously  until  a  fresh  piece  of  litmus  paper 
turns  red.     This  solution  will  be  neutral  enough  for  our  pur- 
pose, but  if  you  are  very  careful  it  will  be  possible  for  you  to 
show  a  red  and  a  blue  paper  in  the  liquid  at  the  same  time ;  the 
liquid  will  then  be  quite  neutral. 

2.  Filter  the  liquid,  replace  it  in  the  evaporating  dish  and 
boil   it   down   to    dryness.      Heat   the   dish   red   hot    for   five 


-2(5— 

minutes.  Toward  the  end  of  the  evaporation,  if  the  salt  sput- 
ters much,  cover  the  dish  with  a  piece  of  paper.  The  product 
is  common  salt.  Taste  it.  If  you  were  not  particular  to  have 
the  liquid  either  exactly  neutral  or  very  slightly  acid,  it  will 
not  taste  like  good  salt. 

3.  Write  twenty-five  equations  showing  the  formation  of  the 
same  twenty-five  salts  by  this  method.     64-c. 

PREPARATION  OF  SALTS  BY  TREATING  AN  ACID  WITH  A  SALT 
OF  A  MORE  VOLATILE  ACID.     64-d. 

One  acid  is  said  to  be  more  volatile  than  another  if,  when  the 
two  are  heated  together,  the  first  escapes  before  the  second. 

Sulphuric  acid  is  one  of  the  least  volatile  of  acids.  Therefore 
when  it  is  heated  with  almost  any  salt,  the  other  acid  which  that 
salt  contains  is  set  free. 

Many  acids  are  very  unstable  in  the  free  condition,  or  cannot 
exist  in  the  free  condition  at  all,  although  they  form  perfectly 
stable  and  well  known  salts.  When  such  salts  are  treated  with 
sulphuric  acid  we  get  the  decomposition  products  of  the  other 
acid  in  place  of  the  acid  itself. 

These  products  are  generally  water  and  the  acid  anhydride. 
Carbonates  are  salts  of  carbonic  acid.  This  is  an  unstable  acid 
and  cannot  exist  in  the  free  condition.  Therefore  when  a  car- 
bonate is  treated  with  almost  any  acid  we  have  carbon  dioxide 
set  free. 

4.  Dissolve  l/i  t.t.  full  of  sodium  carbonate  in  one  t.t.  full  of 
water  in  beaker,  on  wire  gauze,  on  lamp  stand  with  the  aid  of 
heat.     Decant  into  two  t.t. 

5.  To  one  portion,  in  beaker,  on  gauze,  boiling  hot,  add  dilute 
hydrochloric    acid    little    by    little,    until    effervescence    nearly 
ceases  on  addition.     Boil  down  as  before. 

Repeat  using  dilute  sulphuric  acid  with  other  portion.  Boil 
down  and  crystallize  as  with  sodium  sulphate  in  Ex.  No.  9. 

6.  Write  twenty-five  equations  showing  the  formation  of  the 
same  twenty-five  salts  from  the  carbonates.     64-d. 


—27— 


LABORATORY  EXERCISE  NO.   12. 

THE  FORMATION  OF  SALTS  BY  THE  MIXING  OF  TWO  SOLUTIONS 
THAT  COMBINE  TO  FORM  AN  INSOLUBLE  COMPOUND :  ALSO 
BY  FUSING  TOGETHER  BASIC  WITH  ACIDIC  OXIDES.  64-C-f. 

Materials.  Barium  chloride  sol. ;  calcium  chloride ;  sodium 
carbonate;  sand;  litharge;  potassium  chromate;  lead  nitrate. 

Apparatus.     Test  tubes ;    sealed  tubes ;    filter  flask. 

Solubility  rule.  All  the  hydrogen,  sodium,  potassium  and 
ammonium  compounds  are  soluble,  also  the  chlorates,  acetates 
and  nitrates  and  all  the  chlorides  except  those  of  silver,  lead  and 
mercurous  mercury.  165,  166. 

1.  Dissolve  separately  about  five  grams  each  of  sodium  car- 
bonate and  calcium  chloride.     Heat  the  solutions  boiling  hot 
and  mix  them  in  a  bottle.     Filter  with  the  suction  pump,  using 
a  clean  filter  flask.     The  filtrate  should  contain  a  slight  excess 
of  sodium  carbonate.     If  the  filtrate  is  not  perfectly  clear,  filter 
through  an  ordinary  filter.      Acidify  the  filtrate  with  hydro- 
chloric acid  and  boil  it  down  to  dryness  in  evaporating  dish. 

When  mixing  reagents  in  a  test  tube  always  add  the 
second  reagent  a  drop  at  a  time  until  you  ascertain  what 
the  result  is  going  to  be,  then  add  more  and  mix  thoroughly. 
It  is  well  also  to  use  only  a  small  quantity  of  the  first  reagent 
and  dilute  it  with  water  if  a  larger  bulk  is  desired. 

Make  the  following  mixtures  in  test  tubes: 

2.  Barium  chloride  with  sulphuric  acid ;  potassium  chromate 
with  barium  chloride;    potassium  chromate  with  lead  nitrate; 
lead  nitrate  with  sulphuric  acid ;  sodium  carbonate  solution  with 
barium  chloride. 

3.  Write    equations    showing    the    formation    of    the    same 
twenty-five  compounds  by  this  method,  making  use  of  silver 
chloride,  barium  sulphate  or  calcium  carbonate  as  the  insoluble 
compound.     64-e. 

4.  Grind  a   very   small   quantity   of   sand   in   mortar  to   an 
impalpable  powder,  mix  it  thoroughly  with  an  equal  bulk  of 
lead  oxide  and  heat  in  a  sealed  tube.     The  sand  is  an  acidic 
oxide,  which  at  a  -high  temperature  unites  readily  with  the  basic 


—28— 

lead  oxide,  forming  a  lead  silicate.     This  is  a  variety  of  glass 
and  is  absorbed  into  the  glass  of  the  sealed  tube. 

5.  Write  equations  showing  the  formation  of  the  silicates, 
SiO3",  chromates,  CrO4",  sulphates,  and  phosphates  of  the  same 
five  positive  radicals,  by  this  method.  64-f. 

LABORATORY  EXERCISE  NO.   13. 

The  Preparation  and  Properties  of  Chlorine. 

Materials.  Manganese  dioxide;  sodium  chloride;  potas- 
sium chlorate ;  chlorine  water ;  red  cloth  ;  paper  with  printing ; 
paper  with  writing  ink  marks;  silver  nitrate. 

Apparatus.     Test  tubes. 

Caution.  When  smelling  unknown  substances,  first  fill  the 
lungs  with  air  and  then  waft  some  of  the  suspected  odor  toward 
the  nose.  Do  not  smell  the  pure  gas. 

In  this  exercise,  as  well  as  all  others  in  which  disagreeable 
odors  are  produced,  the  hoods  should  be  used  as  much  as 
possible. 

Preparation.  Chlorine  is  prepared  in  the  laboratory  by 
removing  the  hydrogen  from  hydrochloric  acid.  That  is  by 
oxidizing  hydrochloric  acid,  and  it  may  be  accomplished  by 
almost  any  oxidizing  agent.  74-76-78. 

1.  To  5/2  s.s.  of  powdered  manganese  dioxide  in  t.t.  add  y* 
inch  of  dil.  hydrochloric  acid,  shake  and  warm  gently.     Note 
color,  odor  and  action  on  moist  litmus  of  the  escaping  gas.     75. 

2.  To  y2  inch  of  cone,  hydrochloric  acid  in  t.t.  add  a  few 
drops  of  cone,  nitric  acid.     Warm  and  note  what  gas  is  given 
off.     77. 

3.  To  y2  inch  of  cone,  hydrochloric  acid  in  t.t.  add  y2  s.s.  of 
potassium  chlorate.     Warm  and  note  the  result.     Is  the  gas 
the  same  in  all  three  of  these  cases? 

4.  Place  in  t.t.  2  s.s.  of  common  salt  and  2  of  manganese 
dioxide.     Mix  them  thoroughly  by  shaking  and  add  four  drops 
of  water;  shake  again.     Then  add  a  little  cone,  sulphuric  acid. 
What  gas  is  given  off?     Is  it  the  same  as  the  gas  that  comes 
from  the  cone,  hydrochloric  acid  bottle? 

5.  Place  in  t.t.  one  l.s.  of  sodium  chloride,   four  drops  of 


water  and  about  three  c.c.  of  cone,  sulphuric  acid.     Note  large 
volumes  of  a  colorless  gas  that  fumes  in  air. 

6.  Assuming  this  gas  to  be  hydrochloric  acid,  what  ought  to 
be  the  effect  of  mixing  manganese  dioxide  with  the  salt  before 
adding  the  sulphuric  acid?     Explain  the  reaction. 

7.  Place  in  chlorine  water  in  t.t.  some  red  cloth,  paper  with 
printing  and  paper  with  writing  ink  marks.     Chlorine  bleaches 
because  it  oxidizes  in  the  presence  of  water.     71. 

8.  To  a  few  drops  of  silver  nitrate  solution  add  some  chlorine 
water,  to  another  portion  add  some  hydrochloric  acid  and  to 
a  third  add  some  sodium  chloride  solution. 

9.  Shake  the  tubes  thoroughly  to  cause  the  precipitate  to 
settle  and  decant  the  excess  of  liquid.     Add  ammonium  hydrox- 
ide to  the  residue.     Then  add  dilute  nitric  acid  carefully  to  the 
ammonium  hydroxide  solution. 

10.  Write  equations  showing  the  formation  of  chlorine  from 
hydrochloric  acid  and  ten  different  oxidizing  agents.     75-76-78. 
Also  write  five  equations  showing  the  preparation  of  chlorine 
from   a   chloride   and   five   different   oxidizing   agents   in   the 
presence  of  sulphuric  acid. 


LABORATORY  EXERCISE  NO.   14. 

PREPARATION  AND  PROPERTIES  OF  HYDROCHLORIC  ACID.     64-d. 

Materials.    Sodium  chloride ;  iron  filings ;  silver  nitrate  sol. 
Apparatus.     Lamp  stand;    flask;   wire  gauze;    funnel  tube; 
glass  bends  ;  bottles ;  rubber  connectors ;  rubber  stoppers. 

1.  Set  up  the  apparatus  as  indicated  in  the  sketch,  fig.  8. 
A   contains   about   20   g.    of    sodium   chloride;    B    is    empty; 
C  contains  water,  but  the  surface  must  not  quite  touch  the 
glass  tube;    D  has  no  rubber  stopper  and  contains  10  c.c.  of 
ammonium  hydroxide  solution.     The  funnel  tube  must  go  to 
the  bottom  of  the  flask. 

2.  Add  8  c.c.  of  water  to  the  flask  through  the  funnel  and 
be  sure  that  it  wets  the  salt  all  through.     Then  add  25  c.c. 
cone,  sulphuric  acid. 

3.  B   becomes   filled   with   HC1   gas   at   once.      The   gas   is 
absorbed  in  the  water  in  C.     What  are  the  white  fumes  in  D? 


—30— 

4.  Place  a  low  flame  under  the  flask,  remove  the  bottle  B, 
and  replace  it  with  another.     Fill  several  bottles  in  this  way 
with  HC1  gas.     Keep  them   right  side  up  and  covered   with 
glass  plates.      In   the  meantime   notice   what   is   going  on   in 
the  water  just  below  the  end  of  the  tube  in  C. 

5.  Draw  out  the  end  of  a  glass  tube  eight  or  nine  inches  long 
to  a  very  small  size  but  be  sure  there  is  a  hole  through  it.     Pass 
the  drawn  out  end  through  one  hole  of  a  two-hole  stopper  and 


Fig.  8. 

stop  up  the  other  hole  with  a  sealed  tube ;  quickly  insert  the 
stopper  in  one  of  the  bottles  filled  with  HC1  gas  and  invert  it 
so  the  long  end  of  the  tube  will  project  downward  into  a  large 
bottle  filled  with  water.  The  water  will  soon  rise  and  produce 
a  fountain  in  the  bottle.  If  it  is  slow  to  start,  allow  some  cold 
water  to  run  over  the  acid  bottle.  See  fig.  9. 

6.  Place  about  5  c.c.  of  strong  ammonium  hydroxide  in  a 
small  bottle,  shake  it  about  and  pour  it  out.     Invert  over  it  a 


covered  bottle  filled  with  HC1  gas.     Withdraw  the  glass  plate 
between  the  two. 

7.  Test  a  bottle  of  the  gas  with  a  burning  splint  to  see  if  it 
supports  combustion. 

8.  Test  the  liquid  in  C  with  litmus  paper.     Note  the  odor  of 
it.     Does  it  smell  like  the  cone,  hydrochloric  acid  in  reagent 
bottle?     Pour  some  of  it  on  iron  filings  in  t.t.  and  prove  that 
the  gas  given  off  is  hydrogen.     Test  a  portion  of  it  with  silver 
nitrate  and  note  that  the  precipitate   dissolves  in   ammonium 
hydroxide.     Heat  some  with  a  little  manganese  dioxide  and  note 
that  chlorine  is  given  off. 


Fig.  9. 

9.  The  white  solid  which  separates  in  the  flask  is  acid  sodium 
sulphate,  HNaSO4.     See  Acid  salt,  Exp.  25. 

10.  Write  equations  showing  the  formation  of  hydrochloric 
acid  from  ten  different  chlorides.     64-d. 


LABORATORY  EXERCISE  NO.   15. 

THE  PREPARATION  AND  PROPERTIES  OF  BROMINE  AND  IODINE. 

75-76. 

Materials.  Potassium  bromide;  potassium  iodide;  starch; 
manganese  dioxide ;  carbon  disulphide ;  bromine  water ;  chlo- 
rine water;  alcohol;  red  calico;  iodine. 

Apparatus.  Test  tubes;  cork  stopper  with  a  hole  in  it; 
medium  glass  bend. 


—32— 

Definitions.  Sublimation.  The  process  of  distilling  a  solid 
in  such  a  way  that  the  vapor  condenses  directly  to  the  solid 
form. 

Sublime.  To  distill  a  solid  in  such  a  way  that  the  vapor 
condenses  directly  to  the  solid  form. 

Sublimate.     The  solid  product  formed  by  sublimation. 

1.  Place  in  a  bulb-test  tube  2  s.s.  of  potassium  bromide,  two 
of  manganese  dioxide  and  ten  drops  of  water. 

2.  Fit  the  tube  with  cork  stopper  and  medium  glass  bend 
as  in  fig.  10.     Fill  a  large  bottle  with  water  and  stand  in  it  a 
t.t.  containing  a  few  c.c.  of  water.     Place  the  end  of  the  glass 
bend  well  down  in  the  test  tube  and  see  that  it  does  not  quite 
touch  the  water. 


Fig.  10. 

3.  Remove  the  cork  stopper  and  add  about  5  c.c.  of  cone, 
sulphuric  acid  to  the  mixture  of  manganese  dioxide  and  potas- 
sium bromide.     Replace   the  stopper  firmly.     The  whole  will 
remain  supported  by  itself. 

4.  Holding  the  burner  in  the  hand,  warm  the  tube  throughout 
its  whole  length  until  five  or  six  drops  of  bromine  have  dis- 
tilled over.     Do  not  inhale  the  vapor  of  bromine.     Do  not 
get  liquid  bromine  on  the  hands. 

5.  Remove  the  tube  from  the  bottle  and  examine  the  drop 
of  bromine.     Fill  the  tube  half  full  of  water  and  shake  thor- 
oughly to  dissolve  the  bromine  and  form  bromine  water.     Use 
this  bromine  water  in  the  following  experiments.     If  you  need 
more  bromine,  use  the  bromine  water  from  the  reagent  bottle. 

Caution.     Carbon  disulphide  is  extremely  inflammable.     Do 
not  have  a  flame  near  when  using  it  in  the  next  experiment. 

6.  Test  the  bleaching  power  of  bromine  water  just  as  you 
did  with  chlorine  water.     Add  some  carbon  disulphide,  about 


—33— 

4  c.c.  to  a  tube  one  quarter  full  of  water  and  shake  it 
thoroughly.  Note  that  the  carbon  disulphide  does  not  dissolve 
in  the  water,  but  ultimately  settles  out  in  a  clear  globule  at 
the  bottom  of  the  tube.  Now  add  some  bromine  water  and 
shake  again.  Bromine  is  soluble  in  carbon  disulphide,  and 
since  the  latter  is  insoluble  in  water,  the  color  of  the  bromine  is 
soon  transferred  to  it,  changing  it  to  a  yellow  or  a  red,  accord- 
ing to  the  quantity  of  bromine  present.  This  is  a  common 
test  for  free  bromine. 

7.  Dissolve  i   s.s.  of  potassium  bromide  in  about  5  c.c.  of 
water  in  a  test  tube,  add  carbon  disulphide  and  shake.     There 
is  no  color  change,  therefore  no  free  bromine.     Now  add  some 
chlorine  water  and  shake  again. 

8.  Fit  a  bulb-test  tube  with  a  cork  stopper  and  straight  glass 
delivery  tube  three  inches  long.      Place  in  the  tube  I   s.s.  of 


Fig.  IT. 

potassium  iodide,  two  of  manganese  dioxide  and  ten  drops  of 
water.  Add  about  5  c.c.  of  cone,  sulphuric  acid.  Replace  the 
stopper  firmly  and  invert  a  dry  test  tube  over  the  delivery 
tube.  See  fig.  n. 

9.  Holding  the  generating  tube  in  the  wire  test  tube  holder, 
heat  it  carefully  at  first  and  finally  boil  the  liquid  to  drive  the 
iodine  up  into  the  inverted  tube.     The  action  is  over  when  the 
purple  color  disappears  from  the  generating  tube. 

10.  Fill  the  receiving  tube  half  full  of  water,  cover  with  the 
thumb  and  shake  thoroughly.     What  does  this  show  you  about 
the  solubility  of  iodine  in  water  ?     The  brown  solution  is  iodine 
water  and  may  be  used  in  some  of  the  following  experiments. 

3 


—34— 

11.  Place  a  small  crystal  of  iodine  (from  the  dish)  in  a  dry 
t.t.  and  heat  it  gently.     Does  it  melt,  boil,  vaporize,  or  sub- 
lime?    Let  the  tube  cool  and  examine  the  inside  carefully.     Fill 
it  half  full  of  water  and  heat  it  to  boiling.     Is  iodine  volatile 
with  steam?     Add  some  carbon  disulphide  and  iodine  water  to 
a  tube  half   full  of  water  and  shake  thoroughly.     This   is   a 
common  test  for  free  iodine. 

12.  Dissolve  YZ  of  a  s.s.  of  potassium  iodide  in  5  c.c.  of  water 
in   test   tube.      Add    carbon    disulphide    and   a    little   chlorine 
water  and  shake  thoroughly.     What  did  the  chlorine  do  to  the 
potassium  iodide?     Repeat,  using  bromine  water. 

13.  Heat  a  t.t.  full  of  water  in  beaker  to  boiling.     Mix  a  s.s. 
of  starch  with  5  c.c.  of  water  in  test  tube  and  pour  the  mixture 
into  the  boiling  water.     This   forms  a  starch  paste   solution. 
To  a  very  small  quantity  of  this  solution  in  a  test  tube  half  full 
of  water  add  a  few  drops  of  iodine  water.     This  is  the  best  test 
for  free  iodine. 

14.  Test  the  solubility  of  iodine  in  alcohol,  also  in  water  con- 
taining a  little  potassium  iodide.     A  solution  of  iodine  in  alcohol 
is  known  as  tincture  of  iodine. 

15.  Pulverize  y2  of  a  s.s.  of  iodine,  place  it  in  test  tube  one- 
quarter  full  of  water  and  pass  in  hydrogen  sulphide  until  all 
the   iodine  has   dissolved.     What  is  the   residue?     71.     Filter 
it  off  and  test  the  filtrate  for  an  acid.     It  contains  hydriodic 
acid. 

1 6.  Write  equations  showing  the  formation  of  bromine  and 
iodine  from  five  different  bromides  and  five  different  iodides  by 
the  use  of  sulphuric  acid  and  manganese  dioxide. 


LABORATORY  EXERCISE  NO.   16. 

SULPHUR. 

Materials.     Sulphur;   carbon  disulphide;   iron  filings. 

Apparatus.  Test  tubes;  evaporating  dish;  combustion 
spoon. 

Note.  Sulphur  is  insoluble  in  water.  Do  not  attempt  to 
wash  with  water,  tubes  that  have  contained  melted  sulphur. 


—35— 

1.  Grind  together  in  the  mortar  about  a  gram  of  sulphur  and 
5  c.c.  of  carbon  disulphide.     Filter  a  few  drops  of  this  through 
a  dry  folded  filter  paper  held  in  the  hand  upon  a  glass  plate. 
After  the  liquid  has  evaporated  examine  the  crystals  under  the 
microscope. 

2.  Spread  a  large  piece  of  paper  on  the  table.     Stand  the 
burner  on  this  in  an  inclined  position.     Fill  a  dry  test  tube  two- 
thirds  full  of  clean  lumps  of  sulphur.     Holding  it  in  the  test 
tube  holder,  heat  it  very  slowly  and  carefully  until  the  sulphur 
is  all  melted  to  a  light  straw  color.     If  it  becomes  dark  colored 
you  have  heated  it  too  fast.     Pour  some  of  this  light-colored 
liquid  into  water.     Now  heat  it  very  much  hotter  and  note  the 
changes  that  it  undergoes.     When  it  begins  to  boil,  pour  it 
rapidly  into  a  large  quantity  of  water.     Remove  the  product 
from  the  water  and  examine  it  at  once.     Keep  this  product  for 
several  days  and  note  the  changes. 

3.  Let  a  tube  half  full  of  melted  sulphur  stand  until  it  begins 
to  solidify  and  crystals  are  seen  lining  the  inside  of  the  tube. 
Then  pour  out  the  liquid  sulphur  from  the  center  of  the  tube, 
Don't  pour  the  liquid  sulphur  into  the  sink. 

4.  Place  a  small  quantity  of  sulphur  in  the  bowl  of  the  com^ 
bustion  spoon,  not  enough  to  fill  it  when  melted,  and  heat  the 
spoon  until  the  sulphur  is  burning  violently.     Lower  the  burn- 
ing sulphur  into  a  bottle.     Keep  the  bottle  covered  with  a  glass 
plate.     Let  the  sulphur  burn  as  long  as  it  will.     What  product 
is  formed  here?     69.     Is  it  a  solid,  a  liquid  or  a  gas?     Is  it 
acidic  or  basic?     Will  it  dissolve  in  water?     What  effect  has  it 
on  litmus  paper? 

5.  Weigh  out  approximately  6  g.  of  iron  filings  and  3  g.  of 
sulphur.       Grind    these    together    thoroughly    in    the    mortar. 
Transfer  the  mixture  to  the  tube  in  which  you  melted  the  sul- 
phur and  from  which  as  much  sulphur  has  been  removed  as 
possible  by  pouring  it  out  in  the  melted  condition.     Heat  the 
bottom  of  the  tube  as  hot  as  possible  in  the  hood  until  the  reac- 
tion begins  and  then  remove  it  from  the  flame  and  observe  the 
progress  of  the  reaction.      The  compound  that  is   formed  is 
ferrous  sulphide.     When  the  tube  is  cool,  break  it  and  remove 
the  solid  lump.     Save  it  for  the  next  experiment. 

6.  Write  ten  equations  showing  the  formation  of  sulphides 
by  the  direct  union  of  sulphur  with  metals.     40. 


-36- 


LABORATORY  EXERCISE  NO.  17. 

PREPARATION  AND  PROPERTIES  OF  HYDROGEN   SULPHIDE  AND 
SULPHUR  DIOXIDE. 

Materials.  Ferrous  sulphide,  from  last  experiment.  Cu- 
.  pric  sulphate  sol. ;  barium  chloride  sol. ;  lead  nitrate  sol. ; 
potassium  permanganate  sol. ;  red  cloth ;  potassium  bichro- 
mate sol. ;  cadmium  sulphate ;  cylinder  of  liquid  sulphur  dioxide. 

Apparatus.  Bottles ;  beaker ;  evaporating  dish ;  test  tubes ; 
stopper;  delivery  tube. 

As  far  as  possible  all  experiments  with  hydrogen  sulphide 
and  with  sulphur  dioxide  must  be  performed  in  the  hoods. 

1.  Treat  the  lump  of  ferrous  sulphide  obtained  in  the  last 
experiment  in  a  t.t.  with  dilute  sulphuric  acid  and  warm  to 
hasten  the  reaction.     Fit  the  tube  with  a  rubber  stopper  and 
delivery  tube  and  collect  some  of  the  gas  in  a  test  tube  by  dis- 
placement of  air.     Ignite  the  gas.     What  are  the  products? 
68.     Represent  by  equation. 

2.  Pass  some  hydrogen  sulphide  from  the  supply  bottle  in 
the  hood  into  a  test  tube  half  full  of  water  containing  a  little 
cupric  sulphate   solution.     Filter  off  the  precipitate  and  pass 
the  gas  again  into  the  nitrate.     Ascertain  whether  all  the  copper 
can  be  removed  in  this  way. 

3.  Repeat,  using  lead  nitrate  solution.     Also  cadmium  sul- 
phate solution,  formula,  CdSO4. 

4.  Test  the  action  of  a  water  solution  of  hydrogen  sulphide 
on  litmus  paper.      What  would  be   formed  if  hydrogen   sul- 
phide   were    passed    into    a    solution    of    sodium    hydroxide? 
Ammonium  hydroxide? 

5.  Fill  several  bottles  with  sulphur  dioxide  from  the  cylinder 
in  the  hood.     Keep  them  right  side  up  and  covered  with  glass 
plates.     Be  sure  to  turn  off  the  gas  at  the  cylinder. 

Invert  one  of  the  bottles  in  a  dish  of  water.  Fill  another 
bottle  one-quarter  full  of  water,  cover  tightly  with  the  palm 
of  the  hand  and  shake  thoroughly;  note  the  suction  on  the 
hand.  What  does  this  indicate?  Test  the  liquid  with  litmus 


—37— 

paper.     What  acid  is  in  the  water?     Represent  its  formation 
by  an  equation.     Save  the  solution. 

6.  Add  a   few  drops  of  cone,   nitric  acid  to  a  bottle  con- 
taining sulphur  dioxide.     73.     Pour  some  potassium  perman- 
ganate solution  into  some  sulphur  dioxide  water.     Also  some 
potassium    bichromate    solution.       These    are    both    oxidizing 
agents   and   change   the   sulphur   dioxide   to   sulphur   trioxide, 
which  remains  dissolved  in  the  water  in  the  form  of  sulphuric 
acid.     Indicate  these  reactions  by  equations.     75-76. 

7.  Neutralize    some    sulphur    dioxide    water    with    sodium 
hydroxide  solution  and  evaporate  it  to  dryness  in  the  porcelain 
dish.     64-g.     The  product  is  sodium  sulphite,  Na2SO3.     If  this 
be  treated  with  an  acid,  sulphur  dioxide  will  be  given  off. 

8.  Represent  the  formation  of  hydrogen  sulphide  from  five 
different  sulphides  by  the  action  of  acids. 

Represent  the  formation  of  sulphur  dioxide  from  sulphuric 
acid  by  the  action  of  five  different  metals.  71. 

Represent  the  oxidation  of  hydrogen  sulphide  to  sulphuric 
acid  by  five  different  oxidizing  agents.  78. 

Represent  the  oxidation  of  sulphur  dioxide  to  sulphuric  acid 
by  five  different  oxidizing  agents. 


LABORATORY  EXERCISE  NO.   18. 

PREPARATION  AND  PROPERTIES  OF  SULPHURIC  ACID. 

Materials.  Sulphur ;  barium  chloride  sol. ;  bromine  water ; 
cupric  sulphate  sol. ;  zinc  sulphate ;  iron  wire ;  potassium  per- 
manganate sol. 

Apparatus.     Combustion  spoon  ;   test  tubes  ;   bottles. 
Caution.     Do  not  spill  sulphuric  acid. 

Do  not  pour  water  into  sulphuric  acid. 

Do  not  pour  boiling  sulphuric  acid  into  water. 

i.  Place  about  l/2  inch  of  water  in  a  large  bottle.  Place  a 
small  quantity  of  sulphur  in  the  combustion  spoon,  less  than 
will  fill  it  when  melted,  and  heat  it  until  it  is  all  melted  and 
blazing.  Suspend  the  blazing  sulphur  directly  over  the  water, 
keeping  the  bottle  covered,  until  it  will  burn  no  longer.  Remove 


—38— 

the  spoon,  cover  the  bottle  with  the  hand  and  shake  thoroughly 
to  dissolve  the  sulphur  dioxide  in  the  water.  This  forms  a 
solution  of  sulphurous  acid.  To  a  portion  of  this  solution  add 
a  few  drops  of  barium  chloride  solution.  What  does  the 
slight  white  precipitate  indicate?  Then  add  a  few  drops  of 
cone,  nitric  acid.  What  does  the  additional  white  precipitate 
indicate?  73. 

2.  Repeat,  using  bromine  water  in  place  of  nitric  acid.     71. 

3.  Repeat,  using  potassium  permanganate   sol.     76.     Write 
all  equations. 

4.  Notice  the  weight  of  the  cone,  sulphuric  acid  in  the  reagent 
bottle,   also  its  oily  nature.     Pour  some  cone,   sulphuric  acid 
into  10  c.c.  of  water  in  a  test  tube.     Put  pieces  of  wood  and 
paper  in  hot  cone,  sulphuric  acid.     Dilute  some  sulphuric  acid 
with  ten  volumes  of  water  and  write  on  paper  with  it.     Heat 
the  paper. 

5.  Ascertain  the  effect  of  cold  and  hot  cone,  sulphuric  acid 
on  iron  wire. 

Hot  cone,  sulphuric  acid  is  an  oxidizing  agent  according  to 
the  equation,     H2SO4  =  H2O  +  SO2  +  O. 
O  +  Fe  =  FeO. 
FeO  +  H2S04  =  FeS04 .+  H2O. 

2H2SO4  +  Fe  =  FeSO4  +  SO2  +  2H2O. 

What  is  the  effect  of  hot  dilute  sulphuric  acid  on  iron  wire? 

6.  Systematically  dilute  some  of  the  acid  in  ten  test  tubes. 
Find  approximately  how  dilute  a  solution  will  give  a  test  with 
barium  chloride. 

7.  Test  solutions  of  cupric  sulphate  and  zinc  sulphate  with 
barium  chloride. 

8.  Test  for  sulphuric  acid  or  a  soluble  sulphate.      Add 
barium  chloride  and  some  hydrochloric  acid  to  the  solution  to 
be  tested;   a  white  precipitate  insoluble  in  hydrochloric  acid  is 
evidence  of  the  presence  of  the  sulphate  radical. 

Neutral  and  alkaline  solutions  of  phosphates  give  a  white 
precipitate  of  barium  phosphate,  but  this  is  soluble  in  hydro- 
chloric acid,  hence  the  use  of  acid  in  the  above  test. 

9.  Write  equations  showing  the  oxidation  of  sulphur,  hydro- 
gen  sulphide  and  ammonium   sulphide  to   sulphuric  acid  by 
means  of  nitric  acid.     73. 


—39— 


LABORATORY  EXERCISE  NO.   19. 

THE  PREPARATION  AND  PROPERTIES  OF  NITROGEN  AND 
AMMONIA. 

Materials.  Yellow  phosphorus  in  beads  176 ;  sodium  nitrite ; 
ammonium  chloride;  calcium  oxide;  ammonium  nitrate; 
ammonium  sulphate. 

Apparatus.  Water  pan;  1000  c.c.  bottle;  rubber  stopper 
and  delivery  tube ;  flat  cork. 

i.  Place  a  small  piece  of  yellow  phosphorus  in  the  hollow 
of  a  flat  cork,  float  on  water  in  water  pan  over  shelf,  ignite  the 


Fig.  12. 

phosphorus,  invert  large  bottle  over  it  at  once  and  let  it  stand 
until  all  fumes  have  disappeared.  Why  did  bubbles  of  air  come 
out  of  bottle?  Why  does  water  enter  bottle?  What  proportion 
of  bottle  becomes  filled  with  water? 

The  white  smoke  is  phosphorus  pentoxide,  P2O5. 

The  gas  left  in  the  bottle  is  atmospheric  nitrogen.  Note  the 
lack  of  color,  odor  and  the  fact  that  it  will  not  support  com- 
bustion. 

2.  Chemical  nitrogen  may  be  prepared  by  heating  a  solution 
of  ammonium  nitrite.  As  this  is  not  ordinarily  kept  in  stock, 


—40— 

a  mixture  of  sodium  nitrite  and  ammonium  chloride  is  used. 
Fit  a  test  tube  with  a  rubber  stopper  and  delivery  tube.  See 
fig.  12.  Place  in  it  I  s.s.  of  ammonium  chloride  and  an  equal 
amount  of  sodium  nitrite  with  5  c.c.  of  water.  Heat  the 
mixture  carefully  and  collect  the  gas  over  water.  Indicate  the 
reaction  by  two  equations. 

3.  To  i  s.s.  of  ammonium  chloride  in  t.t.  add  a  small  quantity 
of  sodium  hydroxide  sol.  from  reagent  bottle,  and  warm  gently. 
Note  the  odor  and  effect  of  the  gas  on  moist  litmus  paper ;   on 
a  rod  wet  with  cone,  hydrochloric  acid.     Do  not  dip  the  rod 
into  the  reagent  bottle. 

4.  Repeat,  using  ammonium  nitrate ;   and  again  using  ammo- 
nium sulphate.     The  gas  set  free  is  ammonia,  NH3. 

5.  When  an  ammonium  compound  is  heated  it  breaks  up  into 
ammonia  and  the  corresponding  acid.     137.     As  the  gases  cool 
they  reunite  to  form  the  original  solid  substance. 

Heat  a  gram  or  two  of  ammonium  chloride  in  a  dry  test  tube 
and  insert  in  the  mouth  of  the  tube  a  piece  of  moist  red  litmus 
paper.  Explain  the  color  change  on  the  paper  99.  What 
is  the  reason  for  using  an  alkali  or  an  alkaline  oxide  with  an 
ammonium  compound  to  form  ammonia  ? 

6.  Mix  thoroughly  in  the  mortar  about  5  g.  of  ammonium 
chloride  and  10  g.  of  calcium  oxide.     Transfer  the  mixture  to 
a  test  tube,  adding  only  enough  to  half  fill  it.     Fit  the  tube  with 
a  rubber  stopper  and  medium  bend.     Tap  the  tube  in  a  hori- 
zontal position  to  obtain  an  air  space  over  the  whole  length  of 
the  mixture.     Fix  it  in  the  clamp  on  the  lamp  stand  in  a  nearly 
horizontal  position  with  the  stopper  end  lower,  as  in  fig.  13,  so 
that  any  water  formed  will  not  run  back  on  the  hot  glass.     Heat 
very  carefully,  holding  the  burner  in  the  hand,  but  not  hot 
enough  to  cause  the  flame  to  turn  yellow,  as  a  result  of  melt- 
ing the  tube.     Collect  the  ammonia  in  dry  bottles  by  upward 
displacement  of  air. 

7.  Make  an  ammonia  fountain  exactly  as  you  made  the  hydro- 
chloric acid  fountain  in  Ex.  14,  5. 

8.  Invert  a  bottle  or  tube  of  ammonia  in  a  dish  of  water. 
Drop  some  cone,  hydrochloric  acid  into  a  bottle  of  ammonia. 
Test  ammonia  gas  with  moist  litmus. 

9.  Neutralize  some  ammonium  hydroxide  from  the  reagent 
bottle  with  nitric  acid.     Boil  down  the  solution  in  the  evapo- 


—41— 

rating  dish  until  some  of  it  will  solidify  on  the  end  of  a  glass 
rod.     What  have  you  made?     Save  the  product. 

10.  Write  equations  showing  the  formation  of  ammonia  from 
ten  different  ammonium  compounds,  using  at  least  four  different 
alkalies. 


Fig.  13. 

ii.  Show  by  equations  the  formation  of  ten  different  ammo- 
nium compounds  by  the  direct  union  of  ammonia  with  ten 
different  acids. 


LABORATORY  EXERCISE  NO.  20. 

SOME  OXIDES  OF  NITROGEN  AND  NITRIC  ACID. 

Materials.  Ammonium  nitrate ;  scrap  copper ;  sodium 
nitrite ;  phosphorus ;  sodium  nitrate. 

Apparatus.  Lamp  stand ;  water  pan ;  bottles ;  rubber  stop- 
per and  delivery  tube;  funnel  tube. 

i.  Place  in  test  tube  about  il/2  inch  of  ammonium  nitrate. 
Fit  tube  with  stopper  and  delivery  tube.  Fix  tube  in  clamp 
on  lamp  stand  in  an  inclined  position  so  the  bottom  of  tube 


—42— 


comes  near  the  base  of  the  Bunsen  flame.  See  fig.  12.  Heat 
very  carefully  until  the  material  is  all  melted,  then  only  enough 
to  produce  a  steady  evolution  of  the  gas.  Excessive  heating 
may  produce  an  explosion.  Under  no  circumstances  should  the 
material  be  heated  until  it  is  nearly  all  gone.  The  gas  given  off 
is  nitrous  oxide,  N2O.  Remove  delivery  tube  from  water  pan 
before  removing  the  flame. 

2.  Test  the  solubility  of  this  gas  in  water.  Make  the  same 
experiments  with  it  that  you  did  with  oxygen.  How  could  you 
distinguish  this  gas  from  oxygen? 


Fig.  14. 

3.  Fit  a  small  bottle  with  a  two-hole  rubber  stopper. 
Through  one  of  the  holes  pass  a  funnel  tube.  The  end  of  the 
tube  must  reach  very  near  to  the  bottom  of  the  bottle.  Through 
the  other  pass  a  short  bend,  to  which  a  delivery  tube  is  attached. 
Place  several  pieces  of  scrap  copper  in  the  bottle  and  stand  the 
whole  on  the  lamp  stand  with  the  funnel  tube  passing  through 
a  ring  to  prevent  the  apparatus  from  being  upset.  See  fig.  14. 
Arrange  to  collect  the  gas  over  water  in  the  water  pan.  Add 
about  20  c.c.  of  dilute  nitric  acid  and  then  cone,  nitric  acid  little 
by  little  as  necessary  to  produce  the  reaction.  Have  all  receiv- 
ing bottles  upside  down  in  the  pan,  and  absolutely  full  of 


—43— 

water.  Do  not  allow  any  air  to  enter  when  removing  the 
bottles  of  gas  from  the  pan. 

What  is  the  color  of  this  gas?     It  is  called  nitric  oxide,  NO. 

Open  a  bottle  of  it  in  the  air.  Why  was  the  gas  colored 
brown  in  the  generating  bottle  at  first?  Will  it  support  com- 
bustion? Test  it  with  very  hotly  burning  phosphorus  in  the 
hood.  What  would  be  the  effect  of  mixing  nitric  oxide  with 
oxygen?  Nitric  oxide  with  nitrous  oxide? 

4.  Write  ten  equations  showing  the  action  of  nitric  acid  on 
ten  different  metals  when  nitric  oxide  is  formed. 

5.  Treat  sodium  nitrite  with  several  different  acids  in  test 
tube.     Nitrous  acid,  HNO2,  is  unstable  and  cannot  exist  in 


Fig.  15. 

the  free  condition.     Therefore  when  it  is  formed  in  any  reac- 
tion we  get  its  decomposition  products. 

2HNO2  =  H2O  +  N2O3. 

6.  Fill  a  large  bottle  with  water.     Place  in  it  a  test  tube. 
Fit  a  bulb-test  tube  with  a  cork  stopper  through  which  passes 
the  short  arm  of  a  medium  glass  bend.     Place  in  this  tube 
3  s.s.  of  sodium  nitrate,  five  drops  of  water  and  about  5  c.c. 
of  cone,  sulphuric  acid.     Replace  the  stopper  firmly.     See  fig.  15. 

Place  the  long  end  of  medium  glass  bend  in  receiving  tube. 
The  whole  apparatus  will  support  itself.  Holding  the  burner 
in  the  hand,  heat  the  mixture  carefully  until  about  i  c.c.  of 
nitric  acid  has  distilled  over.  Satisfy  yourself  that  this  liquid 
is  nitric  acid  by  its  characteristic  action  on  copper  wire. 

7.  Write  ten  different  equations  showing  the  preparation  of 
nitric  acid  by  the  action  of  sulphuric  acid  on  nitrates. 


—44— 

LABORATORY  EXERCISE  NO.  21. 

EXAMINATION  OF  AN  UNKNOWN  SUBSTANCE. 

The  object  of  this  exercise  is  to  impress  on  the  student  again, 
the  necessity  for  close  original  observation  in  all  his  chemical 
work.  To  call  his  attention  again  to  the  fact  that  every  simple 
experiment  that  he  has  done  or  is  going  to  do  is  simply  a  prac- 
tical way  of  answering  some  question  concerning  the  material 
he  is  operating  on,  and  it  is  for  him  to  see  what  the  question 
is  in  every  case. 

And  that  his  notes  are  simply  a  series  of  definite  facts  which 
he  has  learned  by  his  own  observation  of  the  experiments  done 
in  answer  to  the  various  questions  he  has  had  in  mind. 

Materials.  Several  unmarked  chemical  compounds;  Silver 
nitrate  sol. ;  barium  chloride  sol. ;  potassium  chromate  sol. 

Apparatus.     Sealed  tubes  ;   test  tubes. 

Definitions — Sublime.  A  substance  is  said  to  sublime  when 
it  changes  from  the  solid  to  the  gaseous  form  and  back  to  the 
solid  on  being  distilled.  Water  of  crystallization,  see  lab.  ex. 
No.  8.  Precipitate.  Ex.  No.  8. 

Never  attempt  to  write  equations  unless  there  is  some  definite 
reaction,  the  nature  of  which  is  known. 

1.  Will  it  melt,  boil,  vaporize  or  sublime?     Place  a  small 
quantity  in  a  sealed  tube  and  heat  it  as  ho.t  as  possible,  noting 
results.     Don't  mistake  a  transparent  melted  solid  for  water. 
If  any  water  is  present  it  will  distill  up  on  the  sides  of  the  tube 
or  perhaps  come  out  in  the  form  of  steam. 

Leave  at  least  one  blank  line  after  every  recorded  observation 
where  there  was  a  definite  reaction,  to  be  filled  in  later,  when 
the  nature  of  the  substance  is  known,  with  the  equation  repre-. 
senting  that  reaction. 

2.  Is  it  soluble  in  water,  and  if  so  to  what  extent  ?     Use  very 
small  quantities  in  testing  solubility.     Make  the  test  in  both  cold 
and  boiling  water.     Is  it  soluble  in  dil.  nitric  acid?     How  does 
the  water  solution  affect  litmus  paper? 

3.  How  does  it  affect  the  flame?     Heat  some   on  an   iron 
wire  loop  which  has  been  thoroughly  cleaned  by  dipping  it  in 
hydrochloric  acid  and  washing  it.      Don't  clean  iron  wire  by 
dipping  it  in  the  reagent  bottles. 


—45— 

4.  Treat  portions  of  the  dry  substance,  in  test  tubes,  with 
cone,  sulphuric  acid,  dil.  hydrochloric  acid,  and  sodium  hydrox- 
ide sol.,  warming  in  each  case. 

5.  Test  portions  of  the  water  solution  or  dilute  nitric  acid 
sol.  with  silver  nitrate,  barium  chloride,  potassium  chromate, 
hydrogen  sulphide  gas,  and  sulphuric  acid. 

When  one  unknown  substance  is  finished  exchange  it  for 
another. 

Work  very  slowly  and  be  sure  of  your  facts  before  recording 
them.  Repeat  tests  several  times  if  necessary. 


LABORATORY  EXERCISE  NO.  22. 

PHOSPHORUS  AND  ARSENIC. 

Materials.  Yellow  phosphorus  in  beads  176;  carbon  disul- 
phide;  red  phosphorus;  hydrogen  disodium  phosphate  sol.; 
silver  nitrate  sol. ;  arsenic  trioxide ;  charcoal ;  metallic  arsenic. 

Apparatus.  Iron  plate;  lamp  stand;  large  dry  bottle;  test 
tubes. 

1.  Dry  a  piece  of  yellow  phosphorus  by  touching  it  to  filter 
paper.     Place  it  on  a  dry  iron  plate  on  ring  of  lamp  stand. 
Ignite  it  and  cover  it  at  once  with  a  large  dry  bottle.     Let  it 
stand  until  it  ceases  to  burn  and  the  white  product  has  settled. 
Scrape   out  the   phosphorus   pentoxide   with   a   wire,   place   it 
on  a  glass  plate  and  allow  one  drop  of  water  to   fall  on  it. 
Breathe  on  some  of  the  product.     Do  you  consider  that  it  has 
a  great  affinity  for  water?     Test  it  with  moist  litmus  paper. 
What  acids  does  phosphorus  form?     Which  one  was  formed 
by  the  union  of  the  pentoxide  with  the  one  drop  of  water? 

2.  Dissolve  one  piece  of  yellow  phosphorus  in  about  4  c.c. 
of   carbon   disulphide   in   t.t.      Pour   the   solution   on   disc   of 
filter  paper  supported  on  ring  of  lamp  stand  in  hood  and  wait 
developments.     Don't  spill  this  solution  on  your  clothes  or 
on  your  hands. 

3.  Dissolve  54  °f  a  s-s-  °f  red  phosphorus  in  about  5  c.c.  of 
cone,  nitric  acid  in  t.t.  with  the  aid  of  heat.     The  nitric  acid 
is  a  strong  oxidizing  agent  and  oxidizes  the  phosphorus  to  the 
pentoxide,  which  dissolves  in  the  water  present  to  form  phos- 


phoric  acid,  H3PO4.  73.  Dilute  the  solution  with  water,  add 
a  few  drops  of  silver  nitrate,  mix  these  two,  then  add  ammonia 
carefully  so  as  not  to  mix.  There  ought  to  be  a  yellow  ring  of 
silver  phosphate  in  the  neutral  zone.  What  does  this  show 
about  the  solubility  of  silver  phosphate  in  both  ammonia  and 
nitric  acid?  To  appreciate  better  the  nature  of  silver  phos- 
phate, add  some  silver  nitrate  to  a  solution  of  hydrogen 
disodium  phosphate. 

4.  Warm  a  little  piece  of  metallic  arsenic  in  a  sealed  tube. 
Let  the  tube  cool.     Shake  out  the  arsenic  and  transfer  it  to 
another  sealed  tube.     Heat  as  hot  as  possible  until  no  further 
change  takes  place.     Note  the  two  modifications  of  arsenic. 

5.  Heat  a  very  minute  piece  of  arsenic  very  carefully  and 
slowly  in  a  dry  test  tube.     Explain  what  takes  place.     Can  any 
crystals  be  seen  in  the  tube  ? 

6.  Heat  a  very  small  quantity  of  arsenic  trioxide  in  a  sealed 
tube.     Place  j£  of  a  s.s.  of  arsenic  trioxide  in  a  sealed  tube  and 
above  it  about  il/2  inch  of  broken  charcoal.     Heat  the  charcoal 
red  hot  first,  and  then  the  arsenic. 

7.  Test  the  solubility  of  arsenic  trioxide  in  acids  and  in  alka- 
lies.    63.     Is  it  basic  or  acidic?     Pass  hydrogen  sulphide  into 
the  hydrochloric  acid  solution. 

8.  Write  equations  showing  the  oxidation  of  phosphorus  and 
arsenic  to  the  corresponding  acids,  each  by  five  different  oxidiz- 
ing agents.    73-75. 


LABORATORY  EXERCISE  NO.  23. 

CARBON. 

Materials.  Powdered  charcoal;  animal  charcoal;  litmus 
solution;  lime  water;  lead  oxide;  soft  coal;  copper  scale; 
solution  of  sulphate  of  quinine;  potassium  dichromate, 
K2Cr207. 

Apparatus.  Lamp  stand;  iron  crucible;  test  tubes;  sealed 
tubes;  beaker. 

i.  Heat  bits  of  wood  and  paper  in  sealed  tubes  and  note  that 
the  volatile  gases  burn.  Also  that  something  like  water  is  given 
off  that  has  an  acid  reaction  toward  litmus  paper.  What  is 
left  in  the  tube  in  each  case? 


—47— 

2.  Fill  iron  crucible  half  full  of  animal  charcoal,  cover  with 
iron  plate,  and  heat  it  red  hot  for  five  minutes.     While  this  is 
being  heated  make  a  solution  of  hydrogen  sulphide  by  passing 
the  gas  into  water  in  a  test  tube.     Add  5  c.c.  of  the  charcoal 
to  5  c.c.  of  the  hydrogen  sulphide  solution.     Shake  thoroughly 
and  filter.     Note  that  the  odor  has  been  entirely  removed  from 
the  water. 

3.  Boil  5  c.c.  of  the  freshly  ignited  charcoal  with  5  c.c.  of 
litmus  solution  for  two  minutes  and  filter. 

4.  Shake  a  solution  of  quinine  with  some  of  the  charcoal, 
filter  and  taste  the  filtrate. 

5.  Shake  some  potassium  bichromate  solution  with  charcoal 
and  filter  through  a  dry  filter. 

6.  What  classes  of  bodies  does  charcoal  remove  from  solution  ? 

7.  Place  in  a  sealed  tube  an  intimate  mixture  of  lead  oxide 
and  powdered  wood  charcoal,  connect  the  tube  to  a  medium 
glass  bend,  heat  as  hot  as  possible  and  pass  the  evolved  gas  into 
lime  water.    Is  there  any  evidence  of  lead  in  the  tube  ?    80. 

8.  Repeat  with  copper  scale. 

9.  Place  some  soft  coal  in  a  sealed  tube  and  heat  as  hot  as 
possible.     What  is  the  product  left  in  the  tube? 

10.  Write  ten  equations  showing  the  reduction  of  ten  common 
metals  from  their  oxides.     82. 


LABORATORY  EXERCISE  NO.  24. 

THE  PREPARATION  AND  PROPERTIES  OF  CARBON  DIOXIDE. 

Materials.  Broken  marble ;  charcoal  in  lumps  ;  wood ;  oxy- 
gen ;  magnesium  ribbon ;  lime  water ;  sodium  carbonate ;  sodium 
bicarbonate;  carbon  dioxide  in  cylinders. 

Apparatus.  Lamp  stand  ;  bottles  ;  funnel  tube ;  wire  gauze ; 
evaporating  dish;  combustion  spoon. 

1.  Fill  a  bottle  with  oxygen  by  displacement  of  air.     Test 
with  splint  to  see  when  it  is  full.       Add  a  little  lime  water. 
Ignite  a  piece  of  charcoal  on  the  combustion  spoon  and  lower 
it  into  the  gas.     What  is  the  white  precipitate? 

2.  Place  several  pieces  of  broken  marble  in  small  bottle,  fit  it 
with  two-holed  stopper,  funnel  tube  and  delivery  tube.     Stand 
it  on  the  lamp  stand  with  the  funnel  tube  passing  through  one 


—48- 

of  the  rings  to  prevent  it  being  upset.  Connect  it  to  a  wash 
bottle  as  indicated  in  fig.  16.  Add  dilute  hydrochloric  acid 
through  the  funnel  tube  and  collect  several  bottles  of  the  gas 
by  displacement  of  air.  Test  with  a  burning  splint  to  see  when 
the  bottles  are  full. 

3.  Holding  a  piece  of  magnesium  wire  in  the  forceps,  ignite 
it  and  thrust  it  into  a  bottle  that  is  partly  filled  with  the  gas. 
What  are  the  black  specks? 

4.  Pour  a  few  drops  of  lime  water  into  a  bottle  containing 


some  of  the  gas.     Blow  through  lime  water  in  wash  bottle. 
What  does  this  show  about  carbon  dioxide  in  the  breath? 

5.  Arrange  the  wash  bottle  so  that  the  suction  pump  will 
suck  air  through  lime  water.     Allow  the  air  to  pass  for  five 
minutes.     What  does  this  show  about  the  amount  of  carbon 
dioxide  in  the  air  compared  to  that  in  the  breath? 

6.  Allow  a  slow  stream  of  the  gas  to  bubble  through  10  c.c. 
of  lime  water  in  a  test  tube.     Note  and  explain  all  changes. 
Boil  the  final  clear  solution. 

7.  Pass  a  stream  of  the  gas  into  a  very  small  quantity  of  sodium 
hydroxide  in  a  test  tube.     What  product  is  formed?     64-g. 

8.  Treat  sodium  carbonate  with  acids  and  represent  the  reac- 
tions by  equations.     64-d. 


—49— 

9.  Heat  some  bicarbonate  of  soda  in  a  test  tube  or  a  sealed 
tube  and  pass  the  gas  into  lime  water. 

10.  Decant  the  liquid  from  the  marble  in  the  generator,  wash 
the  marble  and  return  it  to  the  dish.     Boil  down  the  liquid  or 
a  portion  of  it  in  the  evaporating  dish  over  the  wire  gauze 
until  the  residue  is  perfectly  dry.     What  is  it  ?     Will  it  dissolve 
in  water? 

11.  Write  equations  showing  the  decomposition  of  ten  dif- 
ferent carbonates  by  ten  different  acids.     64-d. 


LABORATORY  EXERCISE  NO.  25. 

ACID  SALTS  AND  BAKING  POWDER. 

Materials.  Carbon  dioxide ;  acid  potassium  tartrate ;  bicar- 
bonate of  soda ;  corn  starch ;  tartaric  acid ;  normal  potassium 
tartrate ;  lime  water. 

Apparatus.  Test  tubes ;  glass  bend ;  rubber  connector ; 
bottles;  beaker;  large  porcelain  mortar. 

Definitions — A  polybasic  acid  is  one  having  more  than  one 
replaceable  hydrogen.  Ex.  H3PO4 ;  H2SO4. 

A  diabasic  acid  is  one  having  two  replaceable  hydrogens. 
Ex.  H2SO4. 

A  tribasic  acid  is  one  having  three  replaceable  hydrogens. 
Ex.  H3PO4. 

An  acid  salt  is  one  formed  by  replacing  part  of  the  hydrogen 
of  a  polybasic  acid  by  a  basic  radical.  HNaSO4 ;  HNa2PO4. 

A  normal  salt  is  formed  by  replacing  all  the  hydrogens  of 
a  polybasic  acid  by  the  same  kind  of  positive  radical.  Ex. 
Na2SO4. 

A  basic  salt  may  be  considered  as  one  formed  by  replacing 
only  part  of  the  oxygen  of  a  basic  oxide  or  part  of  the  hydrox- 
ide of  a  base  by  an  acid  radical. 

i.  Stand  a  test  tube  one-half  full  of  sodium  hydroxide  solu- 
tion in  a  small  bottle  of  water.  By  means  of  'a  medium  glass 
bend  let  a  steady  stream  of  carbon  dioxide  bubble  through  it 
until  a  finely  divided  heavy  white  precipitate  has  formed.  64-g. 

This  will  take  nearly  forty  minutes.  Filter  it  off  by  means  of 
suction,  using  the  filter  flask  and  pump,  set  it  aside  to  dry  and 
4 


satisfy  yourself  at  the  next  laboratory  exercise  that  it  is  bicar- 
bonate of  soda.  Heat  some  in  a  sealed  tube  and  pass  the  gas 
into  lime  water. 

2.  Weigh  out  10  g.  of  normal  potassium  tartrate,  K,,C4H4OG. 
From  the  following  equation,  figure  out  the  proper  quantity  of 
tartaric  acid,  and  weigh  it  out: 

K2C4H406  +  H2C4H406  =  2HKC4H406. 

Dissolve  each  powder  separately  in  50  c.c.  of  hot  water, 
and  pour  the  normal  salt  solution  into  the  acid  solution  in  a 
bottle.  The  heavy  white  precipitate  is  bitartrate  of  potassium, 
acid  potassium  tartrate  or  more  commonly  called  cream  of 
tartar.  When  it  is  cold,  filter  it  off  by  means  of  the  suction 
pump  and  filter  flask,  wash  it  once  or  twice  with  water,  and 
set  it  aside  to  dry.  Taste  it. 

3.  Baking  powders  are  mixtures  of  some  kind  of  a  pulverized 
acid  or  acid  salt  with  bicarbonate  of  soda.      They  generally 
contain  some  starch  to  keep  the  mixture  dry.     The  best  baking 
powder  is  one  made  with  cream  of  tartar.     The  equation  rep- 
resenting its  action  when  moistened  is, 

•HNaCO3  +  HKC4H4O6  =  NaKC4H4O6  +  H2O  +  CO2. 

From  this  equation  figure  out  how  much  bicarbonate  of  soda 
should  be  used  to  go  with  20  grams  of  cream  of  tartar.  Weigh 
out  the  proper  quantity.  Mix  these  two  powders  very  thor- 
oughly in  a  large  porcelain  mortar  or  on  a.  large  sheet  of  paper. 
Add  5  grams  of  corn  starch  and  mix  again.  The  more  thor- 
oughly it  is  mixed  the  better  the  powder  will  be. 

Moisten  a  small  quantity  in  a  test  tube  and  explain  the  reac- 
tion. 

4.  Write  four  equations  for  other  kinds  of  powders  using  the 
following  acids  or  acid  salts:    Tartaric  acid,  acid  potassium 
sulphate,  acid  calcium  phosphate,  HCaPO4,  A12  ( SO4) 3.     Alumi- 
num  sulphate  is   the   commonest  ingredient   of   cheap  baking 
powder.     It  reacts  with  the  bicarbonate  of  soda  to  form  sodium 
sulphate,  aluminum  oxide,  and  carbon  dioxide.     148. 


—SI- 
LABORATORY  EXERCISE  NO.  26. 

SILICON. 

Materials.  Water  glass ;  sodium  wire ;  barium  chloride  sol. ; 
cupric  sulphate  sol. ;  sodium  carbonate ;  lead  oxide ;  hydro- 
fluoric acid ;  sand. 

Apparatus.  Sealed  tubes ;  evaporating  dish ;  mortar ;  square 
of  cloth. 

1.  Heat  a  small  piece  of  sodium  in  a  sealed  tube  until  there  is 
no   further   change.      Glass   is   a   silicate   and   contains   silicon 
dioxide.     Sodium  is  a  powerful  reducing  agent.     When  heated 
with  the  glass  it  takes  oxygen  from  the  silicon  dioxide  and 
liberates  silicon. 

2.  Break  the  tube  by  plunging  it  while  hot  into  water  in  the 
evaporating  dish.     Decant  off  the  excess  of  water.     The  black 
material  is  silicon  adhering  to  the  pieces  of  glass.     To  test  this 
add  sodium  hydroxide  sol.  and  observe  very  closely  to  see  if 
a  colorless  gas  is  given  off.     The  gas  is  hydrogen. 

4NaOH  +  Si  =  Na4SiO4  +  2H2. 

3.  Place  about  10  c.c.  of  water  glass  in  mortar.     Do  not 
measure  it  out.     Add  about  one-half  as  much  dil.  hydrochloric 
acid.     Wash  the  jelly  with  water  and  strain  through  a  piece  of 
cloth.      Transfer  the  material  to  the  iron  plate  or  the  iron 
crucible  and  heat  as  hot  as  possible. 

Na4SiO4  +  4HC1  =  4NaCl  +  H4SiO4  (jelly). 

H4SiO4  heated  =  SiO2  +  2H2O. 

The  white  residue  is  silicon  dioxide  or  silica.  Is  it  an  acidic 
or  a  basic  oxide? 

4.  Test  the  solubility  of  this  material  in  hydrochloric  acid  and 
in  sodium  hydroxide  sol.     64-g. 

5.  Dilute  the  water  glass  with  five  times  as  much  water  and 
divide  into  two  parts.     To  one  add  barium  chloride  solution  and 
to  the  other  cupric  sulphate  sol. 

6.  The  following  equations  represent  the  reaction  involved  in 
the  formation  of  lead  sodium  glass. 

Na2O  +  PbO  +  2SiO2  =  Na2SiO3  +  PbSiO3   or,   since   the 
sodium  oxide  is  usually  added  in  the  form  of  sodium  carbonate, 
Na2CO3  +  PbO  +  2SiO2  =  Na2SiO3  +  PbSiO3  +CO2. 


—52— 

Weigh  out  and  mix  these  things,  using  one  gram  of  the  silica 
you  have  made,  and  the  proper  proportions  of  lead  oxide  and 
sodium  carbonate.  Grind  together  in  mortar  and  heat  in  sealed 
tube.  The  product  is  a  variety  of  glass. 

7.  Warm  a  glass  plate  not  hotter  than  can  be  borne  by  the 
hand.  Cover  one  side  thoroughly  with  beeswax  and  let  the 
plate  cool  and  the  wax  harden  by  itself.  Make  a  drawing  on 
the  wax  with  any  sharp  point,  being  sure  that  the  marks  go 
through  the  wax.  Cut  two  pieces  of  filter  paper  just  the  size 
of  the  plate,  place  them  on  top  of  the  waxed  side  and  have  them 
moistened  with  hydrofluoric  acid.  Leave  the  paper  on  the 
plate  for  five  minutes,  then  remove  it  carefully,  being  particular 
not  to  get  any  of  the  acid  on  the  hands.  Wash  the  plate,  warm 
it  gently  and  remove  the  wax  with  a  towel. 

The  hydrofluoric  acid  has  attacked  the  silica  of  the  glass 
according  to  the  following  equation : 

Si02  +  4HF  =  2H20  +  SiF4. 
The  SiF4  is  a  gas. 


LABORATORY  EXERCISE  NO.  27. 

TIN  AND  LEAD. 

Materials.  Granulated  tin ;  stannous  chloride  solution ;  mer- 
curic chloride  sol. ;  ammonium  sulphide ;  -  strips  of  zinc ;  lead 
wire ;  lead  nitrate  sol. ;  potassium  chromate  sol. 

Apparatus.  Test  tubes;  stopper  and  delivery  tube;  lamp 
stand. 

1.  Dissolve  a  small  piece  of  tin  in  warm  dilute  hydrochloric 
acid.     The  gas  given  off  is  hydrogen.     Fix  the  tube  in  clamp 
of  lamp  stand  with  bottom  of  the  tube  near  the  base  of  the  flame 
as  in  fig.  12,  and  let  the  action  continue  for  at  least  ten  minutes. 
The  solution  contains  stannous  chloride,  SnCl2.     120. 

2.  To  I  c.c.  of  mercuric  chloride  sol.  add  stannous  chloride 
from  reagent  bottle  drop  by  drop,  until  no  further  change  takes 
place.     The  white  precipitate  that  forms  at  first  is  mercurous 
chloride,  HgCl.     The  final  grey  precipitate  is  metallic  mercury. 
The  stannous  chloride  is  changed  to  stannic  chloride,   SnQ4, 
which  remains  in  solution.     Write  two  equations. 


—53— 

3.  Repeat  this  test  with  the  stannous  chloride  which  you 
have  made. 

4.  To  2  c.c.  of  stannous  chloride  solution  add  5  c.c.  of  water, 
and  then  hydrogen  sulphide.     Filter  off  the  precipitate.     Stand 
the  funnel  in  a  clean  test  tube  and  add  a  little  yellow  ammonium 
sulphide  to  the  filter.     Yellow  ammonium  sulphide  contains  an 
excess  of  sulphur  dissolved  in  it.     The  brown  sulphide  dissolves 
because  it  forms  with  the  yellow  ammonium  sulphide  a  soluble 
compound    known    as    ammonium    thiostannate    (NH4)2SnS3. 
This   runs  through  the  filter  into  the  test  tube.      Add  some 
hydrochloric  acid  to  this  solution.     The  following  reaction  takes 
place : 

(NHJ2SnS3  +  2HC1  =  2NH4C1  +  H2S  +  SnS2. 

5.  To  2  c.c.  of  stannous  chloride  solution  add  ten  drops  of 
cone,  nitric  acid,  and  heat  gently.     This  oxidizes  the  stannous 
chloride  solution  to  stannic  chloride.     Represent  by  equation. 

Remember  that  this  solution  contains  an  excess  of  free 
hydrochloric  acid.  77.  Dilute  with  10  c.c.  of  water  and  pass 
in  hydrogen  sulphide.  The  precipitate  is  stannic  sulphide. 
Filter  it  off  and  treat  it  just  as  you  did  the  stannous  sulphide. 

6.  Place   about   5   c.c.   of   stannous   chloride   solution    from 
reagent  bottle  in  t.t.  and  add  a  strip  of  zinc.     121. 

7.  Examine  a  piece  of  lead  wire.     Try  to  dissolve  it  in  sul- 
phuric, hydrochloric,  and  hot  dilute  nitric  acid. 

8.  Test  a  solution  of  lead  nitrate  with  sulphuric  acid  and  with 
potassium  chromate  solution.     Pass  hydrogen  sulphide  into  a 
very  dilute  solution  of  lead  nitrate. 

9.  To  2  c.c.  of  lead  nitrate  solution  add  one  drop  of  sodium 
hydroxide  solution.     Now  add  an  excess  of  sodium  hydroxide. 
Compare  the  corresponding  reaction  with  tin. 

10.  To  5  c.c.  of  lead  nitrate  solution  in  test  tube  add  a  strip 
of  zinc.      Leave  it  for  at  least  ten  minutes.     121. 


-54— 

LABORATORY  EXERCISE  NO.  28 

SODIUM  AND  POTASSIUM. 

Materials.  Sodium  bicarbonate ;  lime  water ;  sodium  car- 
bonate; sodium  chloride;  calcium  chloride;  barium  chloride; 
sodium  nitrate;  sodium  sulphate;  potassium  chloride;  hydro- 
gen potassium  tartrate ;  potassium  nitrate ;  potassium  sulphate ; 
carbon  dioxide. 

Apparatus.  Test  tubes;  stopper  and  delivery  tube;  iron 
wire  with  loop  on  end  ;  glass  plate ;  beaker ;  wire  gauze ;  lamp 
stand ;  copper  wire ;  square  of  blue  glass. 

1.  Shake  up  a  teaspoonful  of  common  salt  in  a  bottle  with 
a  t.t.  full  of  ammonia.     Filter  off  l/2  t.t.  of  the  solution.     Pass 
carbon  dioxide  through  the  solution  for  about  an  hour.     Set 
the  tube  away  until  the  next  laboratory  exercise.     The  precipi- 
tate is  sodium  bicarbonate,  made  by  the  Ammonia,  or  Solvay 
Process.     Filter  it  off  and  prove  that  it  is  bicarbonate. 

2.  Heat  5  c.c.  of  bicarbonate  of  soda  in  a  test  tube  having  a 
delivery  tube  that  passes  into  lime  water.     Do  not  melt  the 
tube.     What  is  left  in  the  tube?     When  the  tube  is  cold  add 
2  or  3  c.c.  of  cold  water.     Explain  the  evolution  of  heat. 

3.  Dissolve  a  very  small  quantity  of  common  salt  in  water 
and  let  some  of  the  solution  evaporate  on  a  glass  plate.     Exam- 
ine the  crystals  under  the  microscope. 

4.  Add   some  calcium   chloride   solution,   also   some  barium 
chloride   solution,   to   separate   portions   of   sodium   carbonate 
solution. 

5.  Clean  an  iron  wire  by  dipping  it  repeatedly  in  a  test  tube 
containing  cone,  hydrochloric  acid,  washing  and  heating,  until 
it  gives  no  color  when  heated  in  the  Bunsen  flame.     Moisten 
it  with  water,  touch  it  to  some  sodium  salt  and  heat  in  the 
Bunsen  flame.     What  color  does  sodium  impart  to  the  flame? 
Observe  this  color  through  the  blue  glass. 

6.  Dissolve  8  g.  of  potassium  chloride  and  10  g.  of  sodium 
nitrate  in  20  c.c.  of  water  in  beaker  on  wire  gauze  on  lamp 
stand,  and  boil  off  half  of  the  water.     Let  the  precipitate  settle 
and  pour  the  liquid  off  into  a  test  tube.     Dissolve  a  very  small 
quantity  of  the  residue  in  water  and  let  a  few  drops  of  the 


—55— 

solution  evaporate  on  a  square  of  glass.  Examine  the  crystals 
under  the  microscope.  What  are  they?  Allow  the  tube  con- 
taining the  liquid  to  stand  until  cold.  Crystals  will  form. 
What  must  they  be? 

7.  Test  compounds  of  potassium  in  the  flame  just  as  you  did 
the  sodium.  Note  the  color  through  the  blue  glass.  Put  both 
compounds  of  sodium  and  potassium  on  the  wire  at  once  and 
note  the  flame  color  without  and  with  the  blue  glass.  Which 
color  predominates  without  the  glass?  Which  with  it? 


LABORATORY  EXERCISE  NO.  29. 

CALCIUM,  STRONTIUM,  AND  BARIUM. 

Materials.  Quicklime  in  lumps;  sodium  carbonate;  stron- 
tium nitrate ;  barium  chloride  sol. ;  calcium  chloride. 

Apparatus.  Lamp  stand;  beaker;  wire  gauze;  test  tubes; 
iron  wire ;  flask. 

1.  Heat  3/2  t.t.  of  water  to  boiling  in  beaker  on  wire  gauze. 
Add  a  lump  of  lime  half  the  size  of  an  egg.     The  lime  ought 
to  crumble  to  a  white,  dry  amorphous  powder.     129.     This  is 
dry  slaked  lime  or  calcium  hydroxide,  Ca(OH)2. 

2.  Add  water  and  stir  to  the  consistency  of  milk.     This  is 
"  milk  of  lime"  or  whitewash.     Dilute  a  portion  of  this  very 
much  with  water  and  filter.     Feel  the  filtrate  with  the  fingers, 
test  it  with  litmus  paper  and  taste  it.     It  is  a  very  dilute  solu- 
tion of  calcium  hydroxide  and  is  known  as  lime  water.     Do  you 
consider  calcium  hydroxide  very  soluble? 

3.  Dissolve  about  4  g.  of  sodium  carbonate  in  boiling  water 
in    flask    and    add    a    tablespoonful    of    milk    of    lime.      Stir 
thoroughly  and  filter.     Feel  the  filtrate  with  the  fingers.     Test 
it  with  litmus  paper.     What  is   it?     What  remained   on   the 
filter? 

4.  Dissolve  some  milk  of  lime  in  hydrochloric  acid  and  test 
the  product  on  iron  wire  in  the  flame.     Add  sodium  carbonate 
solution  to  some  calcium  nitrate  solution. 

5.  Make    some    strontium    nitrate    solution    and    add    some 
sodium  carbonate  solution.     To  another  portion  add  some  sul- 


-56- 

phuric  acid.     Test  some  of  the  dry  salt  moistened  with  hydro- 
chloric acid  in  the  flame. 

6.  Repeat  these  tests  with  barium  chloride.     How  could  you 
distinguish  between  barium  and  strontium? 


LABORATORY  EXERCISE  NO.  30. 

MAGNESIUM,  ZINC,  CADMIUM  AND  MERCURY. 

Materials.  Magnesium  wire;  hydrogen  disodium  phos- 
phate; granulated  zinc;  ammonium  sulphide;  cadmium  sul- 
phate; strips  of  zinc;  mercury;  mercuric  chloride;  hydrogen 
sulphide ;  brass. 

Apparatus.     Bottle ;  evaporating  dish  ;  test  tubes. 

1.  Burn  some  magnesium  wire  in  a  bottle.     Don't  drop  the 
burning  wire  into  the  bottle,  but  hold  it  in  the  forceps  while 
burning.     Shake  the  product  with  water  and  test  with  litmus 
paper.     Is  magnesium  oxide  basic  or  acidic?     What  is  formed 
when  it  is  dissolved  in  water? 

2.  Add   magnesium   wire   to   2   or   3   c.c.   of   dilute  hydro- 
chloric acid  in  porcelain  dish  until  no  more  will  dissolve.     Dilute 
the  solution.     What  does  it  contain  ?     To  a  portion  of  this  solu- 
tion add  ammonium  hydroxide.     What  is  the  precipitate  ?     147. 

3.  To  another  portion  add  10  c.c.  of  hydrochloric  acid  and  an 
excess  of  ammonia.     What  compound  is  present  here  which  pre- 
vents the  precipitation  of  magnesium  hydroxide?     147. 

4.  To  this  same  solution  add  a  few  drops  of  hydrogen  di- 
sodium phosphate.     The  precipitate  is  ammonium  magnesium 
phosphate,  NH4MgPO4.     This  is  the  common  test  for  mag- 
nesium. 

5.  Make  a  careful  test  to  see  if  hydrogen  is  given  off  when 
magnesium  is  boiled  with  water. 

6.  Dissolve  about  a  gram  of  zinc  in  5  c.c.  of  dilute  nitric  acid. 
What  have  you  in   solution?      Represent  the  reaction   by  an 
equation.     What  gas  was  given  off?     Dilute  the  solution. 

7.  To  a  portion  of  this  solution  add  sodium  hydroxide  drop 
by  drop  until  a  precipitate  is  formed.     What  is  the  precipitate? 
Then  add  an  excess  of  the  hydroxide.     Why  does  the  precipitate 
dissolve?     Toward  a  strong  acid,  zinc  hydroxide  acts  like  a 
base;    toward  a  strong  base  it  acts  like  an  acid,  having  the 


—57— 

formula  H2ZnO2.  What  soluble  compound  is  formed  when 
zinc  hydroxide  dissolves  in  sodium  hydroxide?  What  other 
hydroxides  act  like  zinc  hydroxide  in  this  respect? 

8.  Pass  hydrogen  sulphide  into  10  c.c.  of  ammonia  in  a  test 
tube    for   five   minutes.      142.      What   compound   is    formed? 
Add  this  solution  to  a  small  quantity  of  the  zinc  solution.     What 
is  the  precipitate?     Is  this  precipitate  soluble  in  dilute  hydro- 
chloric acid?     144.     Is  copper  sulphide  soluble  in  dilute  hydro- 
chloric acid?     How  could  you  separate  zinc  and  copper?     145. 
Dissolve  a  very  small  piece  of  brass  and  prove  that  it  contains 
both  zinc  and  copper. 

9.  Find  out  what  cadmium  sulphide  is  like  and  whether  it  is 
soluble  in  dilute  acids. 

10.  Dissolve  a  very  small  globule  of  mercury  in  2  or  3  c.c. 
of  hot  dilute  nitric  acid.     Dilute  the  solution.     Pass  hydrogen 
sulphide  into  a  portion  of  the  solution  and  dip  a  bright  piece 
of  copper  wire  into  another  portion.     121.     The  solution  con- 
tains mercuric  nitrate. 

11.  Cover  a  small  globule  of  mercury  with  cold  dilute  nitric 
acid  and  let  it  stand  until  the  next  laboratory  exercise.     This 
solution  will  contain  mercurous  nitrate,  HgNO3.     To  a  portion 
of   it  add  a    few   drops   of   hydrochloric  acid.      What   other 
chlorides   are   insoluble?     Filter   off  the  precipitate   and   add 
ammonia  to  the  filter. 


LABORATORY  EXERCISE  NO.  31. 

COPPER  AND  SILVER. 

Materials.  Copper  wire ;  silver  wire ;  iron  wire ;  blue  vitriol ; 
grape  sugar  solution ;  cupric  sulphate  sol. ;  potassium  iodide 
sol. ;  silver  nitrate  sol. ;  sodium  chloride. 

Apparatus.    Test  tubes. 

I.  Place  about  one  inch  of  dilute  nitric  acid  in  a  t.t.  and  add 
as  much  copper  wire  as  the  acid  will  dissolve.  Do  this  in 
the  hood.  Describe  and  explain  the  reaction. 

2..  Divide  the  solution  into  two  parts.  Heat  one  of  these  and 
add  ammonia  little  by  little,  with  much  shaking,  until  the  pre- 
cipitate at  first  formed  dissolves  completely.  Stand  the  tube 
in  the  rack  until  the  next  laboratory  exercise.  Beautiful  blue 


-58- 

crystal  01  cupro-ammonium  nitrate  will  be  found.     Formula, 
Cu(N03)24NH8. 

3.  Dilute  the  other  portion  with  water.     Pass  hydrogen  sul- 
phide into  a  portion  of  this  solution  diluted  with  water.     145. 
Pass  hydrogen  sulphide  into  some  very  dilute  cupric  sulphate 
solution.     See  if  it  is  possible  to  remove  all  the  copper  from 
solution  in  this  way.     Heat  some  blue  vitriol  in  a  sealed  tube 
and  note  the  change  in  the  color.     Allow  some  of  the  water 
to  run  back  on  the  anhydrous  material. 

To  5  c.c.  of  grape  sugar  solution  in  test  tube  add  5  c.c.  of 
cupric  sulphate  solution.  Now  add  sodium  hydroxide  solution, 
shaking  until  the  precipitate  first  formed  is  redissolved. 

Warm  carefully,  noting  the  changes.  Let  it  stand.  The 
precipitate  is  cuprous  oxide.  The  grape  sugar  acted  as  a 
reducing  agent:  do  not  attempt  to  make  use  of  its  formula  in 
your  equation". 

4.  Put    a    piece    of    iron    wire   into    some    cupric    sulphate 
solution.     121. 

5.  Add  some  sodium  hydroxide  solution  to  some  cupric  sul- 
phate solution.     Heat  to  boiling.     The  precipitate  first  formed 
is  cupric  hydroxide.     When  it  is  heated  each  molecule  gives 
up  one  molecule  of  water.     What  is  the  black  precipitate  you 
finally  obtain?     Solutions  of  salts  of  many  other  heavy  metals 
behave  in  the  same  way.     See  Lab.  Ex.  35-4. 

6.  To  a  small  portion  of  some  copper  solution  add  ammonium 
hydroxide  little  by  little  at  first,  and  finally  in  excess.     This 
intense  color  is  quite  characteristic  of  copper. 

7.  Dissolve  one  inch  of  silver  wire  in  about  one  inch  of  dilute 
nitric  acid.     Describe  the  reaction  and  write  the  equation.     Add 
water  to  half  fill  the  tube  and  explain  later  why  this  makes  the 
solution  cloudy. 

8.  To  2.  c.c.  of  this  solution  add  a  solution  of  sodium  chloride. 
Boil  the  contents  of  the  tube.     139.     Get  the  precipitate  on  a 
filter  and  expose  it  to  sunlight.     To  another  portion  add  a 
piece  of  copper  wire.     121. 

9.  To  I  c.c.  of  silver  nitrate  solution  add  a  little  potassium 
iodide  solution.     Pass  hydrogen  sulphide  through  a  very  dilute 
solution  of  silver  nitrate. 

10.  Devise  a  method  for  testing  for  silver,  lead  and  mercu- 
rous  mercury  contained  in  the  same  solution., 


—59— 

LABORATORY  EXERCISE  NO.  32. 

IRON  AND  MANGANESE. 

Materials.  Green  vitriol;  fine  iron  wire;  chlorine  water; 
potassium  nitrate ;  red  lead ;  ammonium  sulphate ;  carbon 
dioxide ;  cupric  sulphate  sol. ;  potassium  f  errocyanide  sol. ; 
ammonium  thiocyanate ;  potassium  ferricyanide ;  manganese 
dioxide ;  sodium  nitrite ;  potassium  hydroxide ;  potassium  per- 
manganate; sulphurous  acid. 

Apparatus.  Flask ;  lamp  stand ;  wire  gauze ;  test  tubes ; 
beaker ;  evaporating  dish  ;  iron  crucible. 

1.  Place  40  c.c.  of  water  in  the  flask,  add  10  c.c.  of  cone. 
sulphuric  acid  and  mix  the  liquids  by  shaking.      Weigh  out 
about  10  grams  of  fine  iron  wire  and  place  it  in  the  dilute 
acid.      Heat  the   flask   on   the   lamp    until    the    reaction    will 
proceed  by  itself,  then  stand  the  flask  in  the  hood  until  all 
the  iron  is  dissolved.     Filter  the  hot  solution  and  divide  the 
filtrate  into  two  portions.     Set  one  portion  aside  to  crystallize 
and  dilute  the  other  for  use  in  the  following  experiments. 

2.  Test  the  action  of  the  dilute  and  cone,  cold  and  hot  acids 
on  bits  of  iron  wire.     See  Ex.  18-5. 

3.  To  a  few  drops  of  ferrous  sulphate  sol.  from  I,  add  a 
little   ammonium   hydroxide   sol.      The   precipitate   is    ferrous 
hydroxide.     Watch  it  for  some  time  and  explain  what  happens 
to  it.     Add  cone,  nitric  acid  until  the  precipitate  dissolves,  then 
boil.     Cool,  and  again  make  alkaline  with  ammonia.     What  did 
the  nitric  acid  do  to  the  ferrous  sulphate  solution?     77. 

4.  Make  a  little  ferrous  chloride  solution  by  dissolving  iron 
wire  in  hydrochloric  acid,  and  add  chlorine  water  in  excess. 
An  excess  is  present  if  an  odor  of  chlorine  remains  after  warm- 
ing.     Add  ammonium  hydroxide  and  explain  result.      What 
did  the  chlorine  water  do  to  the  ferrous  chloride?     77. 

5.  To  a  few  drops  of   ferric  chloride  sol.  add  ammonium 
hydroxide.      Note  the  color  of  the  precipitate.      It  is   ferric 
hydroxide.     Dissolve  it  in  hydrochloric  acid  and  add  a  bundle 
of  fine  iron  wire.     Heat  to  boiling  from  time  to  time  for  three 
minutes.     Pour  some  of  the  almost  colorless  solution  into  hot 
ammonia.     What  is  the  precipitate?    And  what  did  the  iron 
wire  do  to  the  ferric  chloride?     81. 


6.  Add  a  few  drops  of  ammonium  sulphide  to  a  little  ferrous 
sulphate   solution.      To   a   little    ferric   chloride   solution   add 
ammonium  sulphide  and  acidify  with  hydrochloric  acid.     What 
is  the  undissolved  residue?     Make  the  solution  alkaline  with 
ammonium  hydroxide  and  explain  what  the  ammonium  sulphide 
must  have  done  to  the  ferric  chloride.     82. 

7.  Dissolve  20  g.  of  green  vitriol,  crystallized  ferrous  sul- 
phate, FeSO4  -j-  7H2O,  and  the  proper  quantity  of  ammonium 
sulphate  to  make  ferrous  ammonium  sulphate,  (NH4)2SO4.Fe- 
SO4  +  6H2O,  in  50  c.c.  of  water,  boiling,  in  flask,  and  filter 
into  bottle.      Set  aside  to   crystallize.      How  would  you  test 
these  crystals  for  iron,  ammonium  and  for  sulphate? 

8.  Test  both  ferrous  and  ferric  iron  solutions  with  each  of 
the  following   reagents   and   tabulate  the   results:    Potassium 
ferrocyanide,    potassium    ferricyanide,    and    ammonium    thio- 
cyanate.     Devise  a  method  for  testing  for  ferrous  and  ferric 
iron,  each  in  the  presence  of  the  other. 

9.  Oxidize    a    little    potassium    ferrocyanide    solution    with 
aqua  regia.     77.     Dilute  the  solution  and  add  ferrous  sulphate. 
Explain  the  result. 

10.  What  is  the   action   of   ammonium  hydroxide,   also   of 
ammonium  sulphide,  on  potassium  ferrocyanide  sol.?     Explain. 

11.  Heat  a  small  quantity  of  manganese  dioxide  in  a  sealed 
tube.     What  gas  is  given  off,  and  what  is  left  in  the  tube? 
Heat  34  §"•  of  manganese  dioxide  in  a  t.t.  .with  5  c.c.  of  cone. 
HC1.     What  gas  is  given  off?     What  kind  of  a  reaction  is 
this? 

12.  Shake  up  some  manganese  dioxide  with  a  strong  solu- 
tion of  sulphur  dioxide  in  water.     Add  a  small  piece  of  sodium 
nitrite  to  a  tube  containing  one-fourth  gram   of  manganese 
dioxide  and  5  c.c.  of  dilute  nitric  acid.     Heat  to  boiling.     Add. 
one-half  gram  of  red  lead.     Some  of  the  manganese  is  oxidized 
to  permanganic  acid. 

13.  Fuse  one  gram  of  manganese  dioxide  with  one  gram  of 
potassium  hydroxide  and  a  small  quantity  of  KNO3,  in  iron 
crucible.     Dissolve  out  the  product  with  water,  filter,  and  pass 
carbon  dioxide  into  the  solution.     Potassium  permanganate  is 
formed  in  solution.     77. 

14.  Dilute  some  of  the  ferrous  sulphate  solution  with  water, 
and  add  potassium  permanganate  solution.     77. 


LABORATORY  EXERCISE  NO.  33. 

ALUMINUM  AND  CHROMIUM. 

Materials.  Aluminum  wire  and  foil ;  alum ;  borax ;  sodium 
carbonate;  aluminum  sulphate;  ammonium  sulphate;  ammo- 
nium sulphide;  chrome  alum;  potassium  chromate;  alcohol; 
lead  nitrate ;  barium  chloride ;  silver  nitrate ;  mercuric  chlo- 
ride sol. 

Apparatus.  Lamp  stand;  bottles;  test  tubes;  evaporating 
dish;  beaker;  wire  gauze. 

1.  Test  the   solubility  of  aluminum   in   nitric,   hydrochloric 
and  sulphuric  acids,  both  dilute  and  concentrated,  also  in  sodium 
hydroxide  solution.     Evaporate  the  solution  of  aluminum  chlo- 
ride you  have  made  to  dryness  in  the  hood.     Heat  the  residue. 
When  it  has  cooled  try  to  dissolve  it  in  water. 

2.  Amalgamate  a  strip  of  aluminum  foil  in  mercuric  chloride 
solution,  allowing  the  reaction  to  continue  for  several  minutes. 
Wipe  off  the  solution  with  clean  filter  paper.     121. 

After  a  time  notice  the  remarkable  change  it  has  undergone 
and  feel  it  to  note  the  heat  of  reaction.  68. 

3.  To  an  alum  solution  add  ammonium  hydroxide  solution 
until  sufficient  is  present  to  dissolve  the  precipitate  which  first 
forms.     Boil  a  little  of  this  clear  solution.     The  hydroxide  is 
reprecipitated.      This    is    the    regular   test    for   the   aluminum 
radical  in  solution. 

4.  Put  a  little  alum  solution  in  each  of  three  test  tubes.     Add 
borax  solution  to  one,  sodium  carbonate  solution  to  another  and 
sodium    hydroxide    to    the    third.      Compare    the    results    and 
explain.     148. 

5.  Take    10    g.    of    aluminum    sulphate    and    the    required 
quantity  of  ammonium  sulphate  to  make  ammonium  alum.    Dis- 
solve the  two  salts  together  in  50  c.c.  of  boiling  water.     Filter 
the  solution  into  a  bottle  and  set  aside  to  crystallize.     After 
several  days  crystals  of  ammonium  alum  will  be  found.     Form- 
ula, (NH4)2SO4 .  A12(SO4)3  +  24H2O.    Get  these  out,  dissolve 
them  in  water  and  test  the  solution  for  ammonium,  sulphate  and 
aluminum  radicals. 

6.  Add  a  few  drops  of  ammonium  sulphide  to  an  alum  solu- 
tion.    Filter  off  the  precipitate  and  test  it  for  a  sulphide.     148. 


7.  Add  ammonium  hydroxide  to  a  little  chrome  alum  solution. 
Compare  the  precipitate  with  that  formed  by  treating  common 
alum  in  the  same  way.     Treat  another  portion  of  chrome  alum 
solution  with  sodium  hydroxide  and  observe  its  action  with  an 
excess  of  the  precipitant. 

8.  Precipitate  chromic  hydroxide  with  sodium  hydroxide  and 
then  add   enough  of   the   latter   to  redissolve   the   precipitate. 
Dilute  the  solution  with  an  equal  volume  of  water  and  then  pour 
away  all  but  about  an  inch  in  a  test  tube  of  the  solution.     Add 
about  5  g.  of  chloride  of  lime  and  boil  for  several  minutes 
until  the  solution  is  quite  yellow.     Filter  into  a  test  tube  and 
acidify  the  filtrate  with  hydrochloric  acid.     Note  and  explain 
the  successive  color  changes. 

9.  Dissolve  10  g.  of  potassium  bichromate  in  50  c.c.  of  water 
in  flask.     When  solution  is  complete  add  10  c.c.  of  cone,  sul- 
phuric acid.     Transfer  a  few  drops  to  a  test  tube  and  add 
ammonia.    To  the  rest  of  the  solution,  under  the  hood,  add  5  c.c. 
of   alcohol.     The   mixture    of    sulphuric   acid   and   potassium 
bichromate  oxidizes  the  alcohol  to  aldehyde   (C2H4O),  which 
has  a  strong  irritating  odor.     Add  ammonia  to  a  few  drops  of 
this  solution  in  a  test  tube.     Leave  the  rest  of  the  solution  in  a 
bottle  to  crystallize. 

10.  Test   potassium   bichromate   solution   with   lead   nitrate, 
barium  chloride  and  with  silver  nitrate. 


LABORATORY  EXERCISE  NO.  34. 
Review  Laboratory  Work. 

I.     The  Identification  of  Common  Chemical  Compounds. 
Preliminary  Examination. 

1.  If  it  is  a  solid,  pulverize  it  and  put  it  in  a  bottle  or  tube 

marked  with  its  letter  or  number. 

2.  If  you  recognize  it  or  think  you  do,  make  the  final  con- 

firmatory tests  as  called  for  in  these  directions  at  once. 

3.  Test  its  action  toward  moist  litmus  paper. 

a.  Turns  red,  shows  an  acid  anhydride,  acid  or  acid  salt. 

Or  salt  of  a  strong  acid  with  weak  base. 

b.  Turns  blue.     Shows  an  alkali,  an  alkaline  oxide  or  an 

alkaline  carbonate  or  alkaline  salt  of  weak  acid. 


-63- 

4-  Heat  a  small  portion  in  a  sealed  tube  as  hot  as  possible. 

a.  There  may  be  no  change,  possibly  a  sulphate. 

b.  It  may  give  off  water,  showing  water  of  crystallization 

or  water  of  decomposition. 

c.  It  may  give  off  a  colorless  gas  which  turns  lime  water 

white,  showing  a  carbonate  or  bicarbonate. 

d.  It  may  give  off  red  fumes,  showing  a  nitrate  or  nitrite. 

e.  It  may  leave  a  black  residue  of  carbon,  showing  it  to 

be  an  organic  compound. 

/.  It  may  simply  melt,  showing  nothing. 

g.  It  may  give  a  sublimate  of  sulphur,  showing  it  to  be  a 
thiosulphate. 

h.  It  may  volatilize  completely,  showing  it  to  be  either 
ammonium  nitrate  or  ammonium  carbonate. 

i.  It  may  sublime  completely,  showing  it  to  be  an  ammo- 
nium compound  or  a  mercury  compound. 

;'.  The  substance  may  be  yellow  when  hot  and  white  when 
cold,  showing  it  to  be  a  zinc  compound. 

5.  Moisten  a  small  quantity  of  the  solid  with  dil.  HC1. 

a.  Effervescence  of  a  colorless  gas  that  turns  lime-water 

white  shows  a  carbonate  or  bicarbonate. 

b.  Effervescence  of  a  colorless  gas  that  smells  of  sulphur 

dioxide  shows  a  sulphite  or  a  thiosulphate. 

c.  Odor  of  hydrogen  sulphide  shows  presence  of  a  sul- 

phide. 

d.  Odor  of  chlorine  shows  the  substance  to  be  an  oxidiz- 

ing agent. 

6.  Clean  the  iron  wire  loop  by  dipping  it  in  a  t.t.  of  hot  dil. 

hydrochloric  acid,  washing  and  heating  until  it  does  not 
color  the  flame.  Moisten  a  very  small  quantity  of  the 
powder  on  a  glass  plate  with  cone.  HC1.  Place  a  small 
quantity  of  this  mixture  on  the  clean  end  of  the  wire  and 
heat  it  carefully  in  the  edge  of  the  flame. 
a.  Yellow  color  shows  sodium.  Remember  that  sodium 

is   almost  always  present  in   small   quantities   as   an 

impurity. 

Yellow  flames  should  always  be  examined  through 

the  blue  glass.     A  red  color  when  seen  through  the 

blue  glass  shows  potassium. 


-64- 

b.  A  purple  color  appearing  red  through  the  blue  glass 

shows  potassium.     The  blue  glass  removes  the  yellow 
color  due  to  sodium  which  might  obscure  the  purple. 

c.  Red  color  without  the  glass  shows  lithium,  calcium  or 

strontium. 

d.  Green  color  shows  barium  or  copper. 

7.  Heat  about  3  s.s.  of  the  dry  powder  in  a  t.t.  with  3  drops 

of  water  and  3  c.c.  of  cone,  sulphuric  acid. 

a.  No  change ;  possibly  a  sulphate. 

b.  Evolution  of  hydrochloric  acid  gas;   a  chloride. 

c.  Evolution  of  nitric  acid,  a  colorless  liquid  distilling  up 

on  the  sides  of  the  tube,  shows  a  nitrate. 

d.  Evolution  of  sulphur  dioxide  without  the  formation  of 

sulphur  shows  a  sulphite. 

e.  Evolution  of  sulphur    dioxide    with   the    formation    of 

sulphur  shows  a  thiosulphate. 
/.  Bromine  distilling  up  the  sides  of  the  tube  accompanied 

by  acid  fumes  shows  a  bromide. 
g.  A  purple  vapor  shows  an  iodide. 
h.  Evolution  of  an  acid  gas  that  etches  the  inside  of  the 

tube  shows  a  fluoride. 
i.  Odor  of  vinegar  shows  an  acetate. 

8.  Add  a  small  quantity  of  sodium  hydroxide  solution  to  2 

s.s.  of  the  dry  material  in  the  mortar.  Mix  thoroughly 
with  the  pestle.  The  odor  of  ammonia  shows  an  ammo- 
nium compound. 

9.  If  the  material  is  an  acid  or  an  acid  anhydride  it  will  unite 

readily  with  sodium  hydroxide  solution  to  form  a  salt 
with  the  evolution  of  much  heat. 

10.  If  the  material  is  a  base  or  a  basic  oxide  it  will  unite 
readily  with  dilute  nitric  acid  to  form  the  corresponding 
salt  with  the  evolution  of  much  heat. 

The  above  preliminary  tests  should  have  given  some  indica- 
tion as  to  the  nature  of  the  substance.  In  case  they  have, 
proceed  at  once  to  make  the  final  confirmatory  tests  for  the 
radicals  whose  presence  is  suspected.  If  they  have  not,  the 
next  step  is  to  make  a  solution  of  about  5  g.  of  the  material 
either  in  water  or  dilute  nitric  acid.  Make  up  the  solution  to 
loo  c.c.  and  use  small  portions  of  it  in  the  following  tests. 


-65- 

Wherever  a  definite  reaction  is  obtained  leave  a  blank  line  in 
the  note  book  to  be  filled  in  later  with  an  equation  representing 
that  reaction,  when  the  nature  of  the  substance  is  definitely 
known, 

II.     Tests  to  be  made  with  a  Water  or  Nitric  Acid  Solution 
of  the  Material. 

These  tests  must  invariably  be  made  in  the  order  given. 

1.  Add  a  few  drops  of  dil.  HC1  to  a  small  portion  of  the 

solution. 
a.  A  white  ppt.  shows  silver,  lead  or  mercurous  mercury. 

2.  Pass  H2S  through  a  small  quantity  of  the  solution  to  which 

a  few  drops  of  hydrochloric  acid  have  been  added  if  it 
was  not  already  an  acid  solution. 

a.  A  black  ppt.  shows  lead,  copper,  bismuth  or  mercuric 

mercury. 

b.  A  light  yellow  ppt.  shows  cadmium  or  arsenic. 

c.  An  orange  yellow  ppt.  shows  antimony. 

3.  Make  alkaline  with  ammonia,  and  pass  H2S  through  the 

liquid. 

a.  A  black  ppt.  shows  nickel,  cobalt  or  iron. 

b.  A  pink  ppt.  shows  manganese. 

c.  A  white  ppt.  shows  zinc. 

4.  Add  ammonia  and  heat  to  boiling. 

a.  A  white  gelatinous  ppt.  shows  aluminum,  bismuth,  mag- 

nesium, or  a  phosphate. 

b.  A  brown  gelatinous  ppt.  shows  ferric  iron. 

5.  Add  a  solution  of  sodium  carbonate. 

a.  A  white  ppt.  shows  calcium,  strontium,  barium  or  mag- 
nesium. 

6.  To  a  small  portion  of  the  water  solution  add  silver  nitrate. 

a.  A  white  ppt.  shows  chlorine,  nitrite  or  thiosulphate. 

b.  A  cream  colored  ppt.  shows  a  bromide. 

c.  A  light  yellow  ppt.  shows  an  iodide. 

d.  A  bright  yellow  ppt.  shows  a  phosphate. 

e.  A  deep  red  ppt.  shows  a  chromate. 

7.  To  a  nitric  acid  solution  or  solution  to  which  a  few  drops 

of  nitric  acid  have  been  added,  add  silver  nitrate. 
a.  White  ppt.  soluble  in  ammonia  shows  chlorine. 

5 


b.  Cream  colored  ppt.  shows  a  bromide. 

c.  Light  yellow  ppt.  shows  an  iodide. 
8.  Add  barium  chloride  sol. 

a.  White  ppt.  sol.  in  dil.  HC1  shows  a  phosphate  or  car- 

bonate. 

b.  White  ppt.  insol.  in  dil.  HC1  shows  a  sulphate.  ' 

c.  Light  yellow  ppt.  shows  a  chromate. 

III.     Confirmatory  Tests. 

In  each  of  these  tests  use  small  portions  of  a  water  or  nitric 
acid  solution. 

Lead.     With  HC1,  white  ppt.  sol.  in  hot  water ;   with  potass. 

dichromate,  yellow  ppt.;    with  H2S,  black  ppt.;    with 

sheet  zinc,  black  ppt.  of  metallic  lead. 
Silver.     With  HC1,  white  curdy  ppt.  sol.  in  ammonia;  with 

H2S,  black  ppt. ;   with  copper,  white  ppt.  of  metallic  silver. 
Mercurous  mercury.     With  HC1,  white  ppt.  that  is  black- 
ened with  ammonia ;  with  H2S,  black  ppt. ;  covers  bright 

copper  wire  with  mercury. 
Copper.     With  excess   of  ammonia,   deep  blue   sol. ;    H2S, 

black  ppt.;   deposits  copper  on  bright  iron  wire;   boiled 

with   excess   of   sodium  hydroxide  gives  black  ppt.   of 

cupric  oxide. 
Mercuric  mercury.     With  stannous  chloride  gives  at  first  a 

white  ppt.  which  turns  grey  on  adding  an  excess ;   with 

H2S,  black  ppt. ;  deposits  mercury  on  bright  copper  wire. 
Bismuth.     White  gelatinous  ppt.  \vith  ammonium  hydroxide. 

If  this  ppt.  be  collected  on  a  filter  and  a  solution  of 

stannous  chloride  in  sodium  hydroxide  be  added  to  it, 

it  will  be  blackened. 
Cadmium.     With   hydrogen   sulphide   gives    a    yellow    ppt. 

With  sodium  hydroxide  a  white  gelatinous  ppt. 
Antimony.     With  hydrogen  sulphide  an  orange  yellow  ppt. 

of   antimony   trisulphide,    completely   soluble   in   yellow 

ammonium  sulphide. 
Ferrous    Iron.     With    ammonia   a   light   green   ppt.    which 

rapidly  turns  brown  on  the  surface.     With  potass,  ferri- 

cyanide  a  deep  blue  ppt. 


-67- 

Ferric  Iron.  With  potassium  or  ammonium  thiocyanate  a 
deep  red  solution.  With  ammonium  hydroxide  a  brown 
gelatinous  ppt.  With  potass,  ferrocyanide  deep  blue  ppt. 

Aluminum.  With  ammonia,  white  gelatinous  ppt.  soluble  in 
excess  of  ammonia,  insoluble  in  boiling  ammonia. 

Basic  chromium.  When  treated  with  chloride  of  lime,  or 
sodium  oxychloride  and  sodium  hydroxide  it  is  oxidized 
to  chromate  and  recognized  by  its  yellow  color. 

Zinc.  Hydrogen  sulphide  passed  into  ammonia  and  the  solu- 
tion added  to  a  neutral  solution  of  a  zinc  salt  gives  a 
white  precipitate  of  zinc  sulphide. 

Ammonium  hydroxide  gives  a  white  ppt.  of  zinc  hydrox- 
ide easily  soluble  in  an  excess  of  ammonia. 

Nickel.  Sodium  hydroxide  gives  an  apple  green  ppt.  insol- 
uble in  excess  of  sodium  hydroxide. 

Manganese.  Fused  with  a  small  quantity  of  sodium  car- 
bonate and  KNO3  on  platinum  foil  over  blowpipe  gives 
a  green  color. 

Magnesium.  Ammonium  hydroxide  gives  a  white  ppt.  of 
magnesium  hydroxide  soluble  in  excess  of  HC1,  from 
which  solution  it  is  not  reprecipitated  by  ammonia. 
Sodium  phosphate  added  to  this  solution  gives  a  white 
ppt.  of  ammonium  magnesium  phosphate. 

Barium.  Wrhite  ppt.  with  K2SO4  insoluble  in  HC1.  Yellow 
ppt.  with  potass,  chromate.  Green  flame  coloration. 

Strontium.  Red  flame  coloration.  White  ppt.  with  potas- 
sium sulphate. 

Calcium.  Yellowish  red  flame  coloration.  No  ppt.  in  very 
dilute  sol.  with  K2SO4 ;  white  ppt.  with  Na2CO3  sol. 

Sodium.     Bright  yellow  flame  coloration. 

Potassium.  Violet  flame  coloration  which  appears  red 
through  a  blue  glass.  The  sodium  color  obscures  the 
potassium  color,  therefore  all  yellow  colors  must  be 
examined  through  the  blue  glass. 

Chloride.  Silver  nitrate  gives  a  white  curdy  ppt.  insoluble 
in  dil.  nitric  acid  and  soluble  in  ammonia. 

Bromide.  Silver  nitrate  gives  a  cream-colored  ppt.  insoluble 
in  dil.  nitric  acid.  The  original  sol.  treated  with  a  few 
drops  of  cone,  nitric  acid  and  shaken  with  a  little  carbon 
disulphide  gives  a  yellow  or  a  red  globule. 


Iodide.  A  few  drops  of  cone,  nitric  acid  added  to  the  sol. 
give  a  ppt.  of  iodine,  which,  on  boiling,  turns  into  purple 
vapors. 

Sulphide.  The  dry  substance  treated  with  a  little  dil.  sul- 
phuric acid  in  a  t.t.  gives  off  hydrogen  sulphide,  which 
will  blacken  filter  paper  moistened  with  lead  nitrate  and 
ammonia ;  or  the  H2S  gas  passed  into  a  CdSO4  sol.  gives 
a  yellow  ppt. 

Thiosulphate.  The  dry  substance  treated  with  sulphuric 
acid  gives  off  sulphur  dioxide,  and  the  solution  treated 
with  the  acid  gives  a  white  ppt.  of  sulphur  as  well  as 
an  odor  of  sulphur  dioxide.  Silver  nitrate  gives  a  white 
ppt.  of  silver  thiosulphate. 

Sulphite.  Dil.  sulphuric  acid  on  the  dry  substance  gives 
sulphur  dioxide;  with  the  solution  it  gives  an  odor  of 
sulphur  dioxide  without  a  ppt. 

Sulphate.  Barium  chloride  gives  a  white  ppt.  insoluble  in 
dil.  HC1. 

Chromate.     Lead  nitrate  gives  a  yellow  ppt.  of  lead  chromate. 

Nitrite.  Dil.  hydrochloric  acid  gives  an  evolution  of  red 
oxides  of  nitrogen. 

Nitrate.  Concentrated  sulphuric  acid  with  copper  on  the  dry 
salt  gives  nitric  oxide. 

Phosphate.  Silver  nitrate  in  the  neutral  or  water  solution 
gives  a  yellow  ppt.  of  silver  phosphate.  Add  silver 
nitrate  to  the  nitric  acid  solution  and  then  add  ammonia 
carefully.  A  yellow  ring  will  form  in  the  neutral  zone. 

Carbonate.  The  substance  effervesces  with  acids,  giving  off 
carbon  dioxide,  which  will  give  a  white  ppt.  in  lime 
water. 

Acetate.  The  dry  material  heated  with  a  little  cone,  sul- 
phuric acid  and  alcohol  gives  the  odor  of  ethyl  acetate. 

Tartrate.  Add  a  few  drops  of  silver  nitrate  and  then 
ammonia  until  the  ppt.  which  forms  at  first  is  just  redis- 
solved.  Heat  to  boiling  for  some  time.  A  deposit  of 
metallic  silver  in  the  form  of  a  mirror  will  form  .on  the 
inside  of  the  test  tube. 

Bicarbonate.  Gives  water  and  CO2  when  heated  in  sealed 
tube. 


LABORATORY  EXERCISE  NO.  35. 
THE  PREPARATION  OF  CHEMICAL  COMPOUNDS. 

In  this  exercise  the  more  advanced  student  is  expected  to 
select  some  particular  compound  that  he  desires  to  make  on  a 
larger  scale  and  study  out  the  details  of  the  operation  from  the 
following  general  directions.  When  he  knows  exactly  what  he 
ought  to  do  to  obtain  the  result,  he  should  be  provided  with  the 
necessary  materials  and  apparatus. 

GENERAL    DIRECTIONS    FOR    CHEMICAL 
PREPARATIONS. 

Acids.  Oxygen  acids  may  be  made  by  oxidizing  the  required 
non-metal  with  any  suitable  oxidizing  agent  to  the  correspond- 
ing acid  anhydride  and  dissolving  this  in  water. 

Or,  if  they  are  volatile  and  stable,  by  treating  their  salts  with 
sulphuric  acid. 

Basic  oxides. 

1.  By  heating  the  metal  in  oxygen. 

2.  By  heating  the  carbonate. 

3.  By  heating  the  nitrate. 

4.  By  heating  the  hydroxide. 

Salts. 

1.  By  treating  a  metal  with  an  acid. 

2.  By  treating  a  basic  oxide  with  an  acid. 

3.  By  treating  a  base  with  an  acid. 

4.  By  treating  an  acid  with  a  salt  of  a  more  volatile  acid. 

5.  By  double  decomposition  with  the  formation  of  an 

insoluble  compound. 

6.  By  melting  together  a  basic  with  an  acidic  oxide. 

7.  By  treating  an  alkali  with  an  acid  anhydride. 

Preparation  of  salts.  Before  undertaking  the  preparation 
of  a  chemical  compound  of  any  description  it  is  necessary  'to 
look  up  its  properties,  especially  its  solubility,  and  the  best 


method  of  making  it.  If  it  is  desired  to  make  a  chemically  pure 
compound  it  is  desirable  to  use  chemically  pure  materials. 

Insoluble  salts.  Suppose  it  is  desired  to  make  a  compound 
that  is  virtually  insoluble  in  water.  It  is  commonly  made  by 
mixing  two  solutions,  one  of  which  contains  the  positive 
radical  and  the  other  the  negative  radical  of  the  required 
substance. 

Select  two  chemically  pure  and  soluble  compounds  and  cal- 
culate exactly  how  many  grams  of  each  will  be  necessary  to 
form  the  desired  amount  of  the  insoluble  compound,  being  sure 
to  take  into  account  anv  water  of  crystallization  that  they  may 
contain. 

Look  up  their  solubilities  in  water  and  dissolve  each  in  hot 
water  and  add  enough  cold  water  to  make  a  cold  saturated 
solution. 

Filter  the  solutions  and  add  one  solution  to  the  other  very 
slowly  with  violent  stirring. 

Test  a  small  portion  on  a  suction  filter  to  see  if  it  can  be  fil- 
tered and  washed.  If  it  is  found  possible  to  filter  it  in  this  way, 
wash  it  rapidly  by  decantation,  pouring  the  washings  through 
a  large  suction  filter,  finally  transfer  all  the  material  to  the 
filter  and  wash  by  pouring  water  through  it  until  the  washings 
show  no  test  for  the  negative  radical  of  the  other  compound 
formed. 

If  it  is  impossible  to  filter  and  wash  it  on  the  filter  let  it 
settle  in  a  narrow  deep  vessel  over  night  and  siphon  off  the 
wash  water  the  next  day.  Add  more  water,  stir  thoroughly, 
let  settle  and  siphon  again.  Repeat  this  as  many  times  as 
necessary  to  get  rid  of  the  other  compound  present ;  or  it  might 
be  filtered  through  an  ordinary  folded  filter. 

Soluble  salts.  I.  Look  up  the  solubility  of  the  salt  at  100° 
C.  and  at  20°  C.  The  difference  between  these  will  be  approxi- 
mately the  amount  of  the  salt  that  will  crystallize  out  of  100  c.c. 
of  a  hot  saturated  solution  when  it  cools  down  to  20°  C.  If 
it  is  desired  to  have  100  g.  crystallize  out,  as  many  hundred 
c.c.  of  a  hot  saturated  solution  will  be  necessary  as  this  differ- 
ence is  contained  in  100. 

Having  found  the  volume  of  the  saturated  solution,  find  how 
many  grams  of  the  substance  will  have  to  be  made  to  saturate 
it.  From  this  compute  the  quantities  of  the  materials  necessary. 


The  final  solution  is  filtered  hot  and  the  hot  filtrate,  in  a 
beaker  or  flask  well  covered,  is  thoroughly  wrapped  up  in  cloth 
to  cause  it  to  cool  very  slowly,  thereby  forming  large  crystals, 
and  left  to  stand  over  night. 

If  pure  materials  have  been  used  the  product  is  essentially 
chemically  pure.  In  any  case  it  may  be  recrystallized  several 
times  if  necessary  to  render  it  more  pure. 

To  find  the  number  of  cubic  centimeters  of  acid  to  use,  the 
theoretical  weight  of  the  acid  is  divided  by  the  quantity  of  acid 
in  i  c.c.  of  the  solution.  If  much  dilute  acid  is  to  be  used  it 
will  be  necessary  to  take  the  water  it  contains  into  account  in 
computing  the  quantity  of  water  to  add  to  make  up  the  required 
volume. 

Example.  Suppose  it  is  desired  to  make  potassium  nitrate 
and  have  100  g.  crystallize  out  from  solution.  100  c.c.  of 
water  at  100°  C.  dissolve  247  g.  of  the  salt,  and  at  20°  C.,  31  g. ; 
therefore  on  cooling  such  a  saturated  solution  to  20°  C.  216  g. 
ought  to  crystallize  out.  Leaving  out  of  account  the  change  in 
the  volume  due  to  the  presence  of  the  salt,  the  volume  of  water 

which  would  cause  100  g.  to  separate  would  be  — —  X  100  c.c. 

216 

which  equals  46  c.c. 

i  c.c.  of  a  saturated  solution  at  100°  C.  contains  2.47  g.  of  the 
salt.  Therefore  the  total  quantity  to  be  made  in  this  case  is 
2.47  X  46,  which  equals  114  g. 

Suppose  the  salt  is  to  be  made  by  mixing  potassium  carbonate 
with  dilute  nitric  acid.  Then  according  to  the  following 
equation : 

78  g.          71  g-  H4  g- 

K2CO3  +  2HNO3  =  2KNO3  +  H2O  +  CO2. 

138  126  202 

— —  X  114=78  g.  of  potassium  carbonate  and X  114=71 

202  202 

g.  of  nitric  acid  will  be  necessary. 

The  specific  gravity  of  the  acid  is  1.28  and  it  contains  45.5% 
nitric  acid.  One  c.c.  of  it  therefore  contains  .58  g.  of  HNO3 
and  .70  g.  water.  The  number  of  c.c's.  necessary  would  be, 

-^  =24  c.c.     Since  one  c.c.  of  this    acid  contains  .70  c.c.  of 
5° 


—72— 

water  the  whole  will  contain  86  c.c.,  which  is  more  than  that 
required  to  dissolve  the  salt  formed.  No  more  water  need 
therefore  be  added.  The  dilute  acid  is  added  little  by  little  to 
the  carbonate  in  a  flask.  The  heat  of  reaction  will  probably 
be  sufficient  not  only  to  heat  the  solution  to  boiling  but  to 
evaporate  some  of  the  excess  of  water.  129. 

The  preparation  of  the  hydroxides. 

j.  In  the  case  of  the  alkalies,  add  milk  of  lime  to  a  solu- 
tion of  the  carbonate. 

2.  In  the  case  of  the  less  soluble  hydroxides,  add  a  very 
soluble  hydroxide  to  a  solution  of  one  of  the  salts. 

The  preparation  of  acid  salts.  These  can  only  be  prepared 
from  polybasic  acids.  Heat  the  normal  salt  with  the  reouired 
excess  of  acid. 

QUESTIONS   ON  THE   LABORATORY  EXERCISES. 
Exercise  No.  i. 

What  are  the  essential  elements  of  a  Bunsen  burner?  Make 
a  drawing  showing  the  essential  construction.  In  what  four 
ways  may  a  burner  be  found  burning?  Which  of  these  is  the 
proper  way  ?  What  are  the  two  ways  that  are  decidedly 
wrong?  How  may  the  wrong  ways  be  easily  corrected? 
Which  way  is  undesirable  although  harmless?  Describe  the 
flame.  Which  is  the  hottest  part? 

Exercise  No.  2. 

Describe  how  to  cut  glass  tubing.  In  rounding  the  edges  of 
a  glass  tube  why  should  the  other  end  be  held  lower  down? 
Why  does  the  flame  appear  yellow  when  the  glass  is  heated? 
In  making  a  sealed  tube  why  is  it  undesirable  to  leave  a  great 
mass  of  glass  on  the  end?  Why  cannot  glass  tubing  be  prop- 
erly bent  in  the  Bunsen  flame  ? 

Exercise  No.  3. 

Define  the  terms,  acid ;  alkali ;  basic  oxide ;  alkaline  oxide 
and  acidic  oxide.  What  is  the  effect  of  thrusting  a  spark  into 
a  bottle  of  oxygen  ?  If  the  wood  contains  carbon  and  hydrogen, 


—73— 

what  products  would  be  formed?  68.  What  is  the  product 
formed  when  magnesium  burns  in  oxygen?  How  does  this 
moist  product  affect  litmus  paper?  In  what  will  basic  oxides 
generally  dissolve?  State  some  characteristic  properties  of 
acids. 

Exercise  No.  4. 

Define  the  term  "Properties  of  matter/'  What  topics  should 
be  considered  when  describing  the  properties  of  a  given  kind  of 
matter?  See  outline  for  recitation.  Describe  some  proper- 
ties of  sodium.  Describe  the  visible  effects  of  putting  sodium 
on  wet  filter  paper.  Is  sodium  oxide  a  basic  or  acidic  oxide? 
What  would  be  formed  if  it  were  dissolved  in  water?  What 
is  the  effect  of  heating  mercuric  oxide?  What  gas  comes  out 
of  the  tube  ?  What  is  left  in  the  tube  ?  What  is  left  in  the  tube 
after  heating  maganese  dioxide?  71.  What  is  the  effect  of 
heating  red  oxide  of  lead?  What  kinds  of  oxides  dissolve 
readily  in  acids? 

Exercise  No.  5. 

Why  is  a  mixture  of  manganese  dioxide  and  potassium 
chlorate  used  for  preparing  oxygen?  What  is  left  in  the  tube 
after  no  more  oxygen  can  be  obtained?  How  could  you  prove 
this?  Why  does  the  oxygen  look  smoky?  What  compound 
is  supposed  to  be  formed  when  carbon  dioxide  dissolves  in 
water?  What  kind  of  an  oxide  is  CO2,  acidic  or  basic? 
What  two  oxides  does  phosphorus  form?  Under  what  condi- 
tions would  each  result?  What  acids  would  they  form  with 
water?  What  oxides  does  sulphur  form?  What  acids  would 
these  oxides  form  with  water?  What  gases  would  burn  in 
oxygen  ?  Would  oxygen  burn  in  any  gas  ? 

Exercise  No.  6. 

Why  should  the  bottles  be  kept  inverted?  What  would  be 
the  color  of  the  flame  if  the  gas  issued  from  a  platinum  tube? 
Why  is  it  yellow  when  burning  on  the  end  of  a  glass  tube?  If 
the  gas  dissolved  in  water,  why  should  there  be  a  suction  on  the 
hand?  What  is  the  most  explosive  mixture  of  hydrogen  and 
air  ?  Why  should  the  hydrogen  coming  directly  from  a  genera- 
tor not  be  ignited  ?  Why  was  water  put  in  the  apparatus  ?  The 


—74— 

action  might  stop  for  three  different  reasons,  what  are  they? 
State  three  ways  in  which  hydrogen  might  be  obtained  from 
water.  Two  ways  in  which  it  might  be  obtained  from  acids. 
When  sulphuric  acid  acts  on  zinc,  what  else  is  formed? 
Describe  three  different  experiments  proving  that  hydrogen  is 
lighter  than  air.  Describe  one  experiment  proving  that  water 
is  formed  when  hydrogen  burns. 

Exercise  No.  7. 

What  is  meant  by  electrolysis?  What  is  an  electrolyte? 
Which  is  the  positive  electrode?  Which  the  negative?  Will 
distilled  water  allow  the  current  to  pass  through  it?  Will 
hydrant  water  allow  the  current  to  pass?  What  effect  has  the 
presence  of  a  small  quantity  of  sulphuric  acid  on  the  conduc- 
tivity of  distilled  water?  What  are  the  visible  effects  of  pass- 
ing a  current  through  a  solution  of  potassium  sulphate  colored 
with  litmus  solution? 

Exercise  No.  8. 

What  is  meant  by  reduction?  What  is  a  reducing  agent? 
What  is  a  precipitate?  When  is  a  substance  said  to  be  crystal- 
line? When  amorphous?  What  is  water  of  crystallization? 
What  is  meant  by  efflorescence?  What  is  meant  by  distilla- 
tion and  what  is  the  distillate?  What  is  the  effect  of  heating 
a  metallic  oxide  in  an  atmosphere  of  hydrogen?  What  class  of 
impurities  does  filtration  remove  from  water?  What  kinds  of 
impurities  does  distillation  remove? 

Exercise  No.  9. 

What  is  meant  by  a  cold  saturated  solution ;  a  hot  saturated 
solution ;  a  super-saturated  solution  ?  How  are  such  solutions 
made?  How  may  a  super-saturated  solution  be  caused  to 
crystallize?  What  is  water  of  crystallization?  What  is  the 
complete  formula  of  Glauber's  Salt? 

Exercise  No.  10. 

What  is  an  acid;  a  base;  a  salt;  a  basic  oxide?  What  is 
the  action  of  an  acid  on  a  metal?  What  gas  is  generally 
formed?  Under  what  circumstances  is  this  gas  not  set  free? 


—75— 

How  many  grams  of  hydrochloric  acid  would  be  necessary  to 
exactly  combine  with  ten  grams  of  iron?  151.  What  is  green 
vitriol?  What  is  formed  when  magnesium  dissolves  in  hydro- 
chloric acid?  What  is  the  action  of  an  acid  on  a  basic  oxide? 
What  is  always  formed  in  this  kind  of  action?  How  many 
grams  of  zinc  oxide  would  be  necessary  to  form  ten  grams 
of  zinc  sulphate?  Write  equations  showing  the  formation  of 
any  salt  in  two  different  ways. 

Exercise  No.  n. 

What  is  an  alkali ;  a  base  ?  What  is  the  action  of  an  acid 
on  a  base?  What  is  first  formed?  What  else  is  formed? 
When  is  a  liquid  said  to  be  neutral  ?  What  is  meant  by  decant  ? 
In  i,  just  before  the  solution  became  neutral  there  was  a 
slight  effervescence,  to  what  was  it  due  ?  Why  was  it  desirable 
to  have  the  solution  slightly  acid  rather  than  slightly  alkaline? 
When  is  one  liquid  said  to  be  more  volatile  than  another? 
What  is  the  effect  of  mixing  an  acid  with  a  salt  of  a  more 
volatile  acid?  If  the  more  volatile  acid  is  unstable,  what  pro- 
ducts may  be  given  off?  Name  some  non-volatile  acids;  some 
acids  that  are  volatile  and  stable ;  some  acids  that  are  volatile 
and  unstable.  What  is  the  effect  of  adding  an  acid  to  a  car- 
bonate? How  many  grams  of  sodium  carbonate  would  be 
necessary  to  form  ten  grams  of  sodium  chloride? 

Exercise  No.  12. 

State  the  full  solubility  rule.  166.  What  is  the  effect  of 
mixing  two  solutions  one  of  which  contains  the  basic  radical 
and  the  other  the  acidic  radical  of  an  insoluble  compound? 
How  did  you  make  barium  sulphate?  Describe  lead  chromate, 
calcium  carbonate  and  barium  carbonate.  What  is  formed 
when  a  basic  oxide  is  fused  together  with  an  acidic  oxide?  In 
i,  why  should  the  filtrate  be  acidified  with  hydrochloric  acid? 
Why  should  it  effervesce  on  the  addition  of  the  acid?  If  five 
grams  of  sodium  carbonate  had  been  used,  how  many  grams  of 
calcium  chloride  should  have  been  necessary  to  go  with  it  in 
order  that  there  should  not  be  an  excess  of  either  material? 
Name  the  insoluble  sulphates;  the  insoluble  chlorides.  State 
the  general  conditions  under  which  precipitates  are  formed. 


-76- 

129.  Any  oxygen  salt  may  be  considered  as  made  up  of  what 
two  kinds  of  compounds  ?  Write  equations  showing  the  forma- 
tion of  any  oxygen  salt  in  six  different  ways. 

Exercise  No.  13. 

Represent  all  reactions  involved  in  this  exercise  by  equations. 
How  may  the  hydrogen  be  removed  from  hydrochloric  acid? 
Name  ten  oxidizing  agents  and  state  how  they  oxidize.  How 
many  grams  of  manganese  dioxide  would  be  necessary  to  form 
ten  grams  of  chlorine?  In  2,  why  is  the  color  of  the  escaping 
gas  reddish  brown?  What  acid  should  theoretically  be  formed 
when  hydrochloric  acid  is  added  to  potassium  chlorate?  If 
this  acid  were  volatile  and  unstable,  what  products  might  be 
given  off?  In  a  mixture  of  several  salts,  if  sulphuric  acid  were 
present  in  excess,  what  salts  would  remain  after  heating?  64-d. 
What  kind  of  colors  does  chlorine^  bleach  ?  When  nitric  acid 
is  added  to  ammonium  hydroxide,  what  salt  is  formed?  In  8, 
what  is  the  final  precipitate? 

Exercise  No.  14. 

Why  must  the  funnel  tube  go  to  the  bottom  of  the  flask? 
Why  is  water  put  in  the  flask?  Why  is  hydrogen  chloride 
visible  in  moist  air  ?  What  are  the  white  fumes  in  D  ?  Repre- 
sent all  reactions  involved  in  this  exercise  by  equations.  How 
many  grams  of  hydrogen  chloride  could  be  made  from  twenty 
grams  of  salt?  How  many  liters  would  this  gas  occupy  at 
normal  temperature  and  pressure?  152.  105.  State  all  the 
evidence  you  have  that  the  liquid  in  D  is  the  same  as  the 
hydrochloric  acid  in  the  reagent  bottle.  How  soluble  is  hydro- 
gen chloride  gas  in  water?  In  5,  what  makes  the  water  rise 
into  the  upper  bottle?  How  might  acids  other  than  hydro-, 
chloric  be  made?  Write  equations  showing  the  formation 
of  hydrochloric  acid  from  ten  different  chlorides.  What  would 
be  formed  if  chlorine  were  allowed  to  act  on  wet  phosphorus? 
71,  73 

Exercise  No.  15. 

How  might  bromine  be  obtained  from  hydrobromic  acid? 
What  oxidizing  agents  would  remove  the  hydrogen  from 
hydrobromic  acid?  Is  hydrobromic  acid  more  or  less  stable 


—77— 

than  hydrochloric  acid  ?  What  takes  place  when  chlorine  water 
is  added  to  a  solution  of  a  bromide?  Is  hydriodic  acid  more  or 
less  stable  than  hydrobromic  acid  ?  Could  this  acid  be  made  by 
the  action  of  sulphuric  acid  on  an  iodide?  Is  iodine  volatile 
with  steam?  How  may  iodine  be  separated  from  an  iodide? 
What  is  the  effect  of  adding  either  chlorine  or  bromine  water 
to  a  solution  of  an  iodide?  Describe  tests  for  both  free  and 
combined  iodine.  What  is  the  effect  of  passing  hydrogen 
sulphide  into  iodine  suspended  in  water?  What  would  be 
formed  if  phosphorus  were  added  to  iodine  water?  Name 
three  good  solvents  for  iodine.  Represent  all  the  reactions 
involved  in  this  exercise  by  equations.  How  many  grams  of 
phosphorus  would  be  necessary  to  form  ten  grams  of  hydriodic 
acid? 

Exercise  No.  16. 

Represent  all  the  reactions  involved  in  this  exercise  by 
equations.  Draw  a  diagram  of  the  crystals  obtained  by  the 
evaporation  of  the  carbon  disulphide  solution.  Describe  all 
the  changes  that  sulphur  undergoes  when  very  slowly  heated. 
Describe  three  different  varieties  of  sulphur.  How  would  you 
change  any  one  of  them  into  any  other?  What  is  formed  when 
sulphur  burns  in  air  ?  What  is  formed  when  this  oxide  is  dis- 
solved in  water?  What  is  the  effect  of  adding  an  oxidizing 
agent  to  this  solution?  How  many  grams  of  ferrous  sulphide 
could  be  made  from  ten  grams  of  iron?  How  many  liters  of 
hydrogen  sulphide  could  be  obtained  from  this  amount  of 
ferrous  sulphide? 

Exercise  No.  17. 

Represent  by  equations  all  the  reactions  involved  in  this 
exercise.  How  would  you  obtain  hydrogen  sulphide  from 
sulphur?  How  many  grams  of  iron  would  be  necessary  to 
produce  ten  liters  of  hydrogen  sulphide  at  normal  temperature 
and  pressure?  What  is  formed  when  hydrogen  sulphide 
burns  ?  What  evidence  have  you  that  hydrogen  sulphide  water 
has  acid  properties  ?  State  three  ways  in  which  sulphur  dioxide 
may  be  made.  How  could  you  make  sodium  sulphite?  Show 
the  formation  of  sulphur  dioxide  by  the  oxidation  of  metals  by 


means  of  sulphuric  acid.  How  many  grams  of  copper  would 
be  necessary  to  form  ten  liters  of  sulphur  dioxide?  Describe 
the  preparation  of  sulphuric  acid.  How  can  you  prove  that 
nitric  acid  changes  sulphurous  acid  to  sulphuric  acid? 

Exercise  No.  18. 

Represent  all  the  reactions  involved  in  this  exercise  by  equa- 
tions. How  many  grams  of  sulphur  would  have  to  be  burned 
to  form  ten  grams  of  sulphuric  acid?  How  many  grams  of 
bromine  would  theoretically  be  necessary  to  oxidize  ten  grams 
of  sulphur  as  sulphurous  acid  to  sulphuric  acid?  Describe  the 
test  for  the  sulphate  radical.  How  would  you  distinguish 
between  a  soluble  phosphate  and  a  sulphate?  Give  the 
formulas  of  ten  oxidizing  agents  and  state  how  they  oxidize. 

Exercise  No.  19. 

Represent  all  the  reactions  of  this  exercise  by  equations. 
What  two  oxides  does  phosphorus  form?  Under  what  cir- 
cumstances is  each  formed?  How  does  burning  phosphorus 
over  water  prepare  nitrogen?  Does  it  really  prepare  it? 
What  becomes  of  the  white  smoke?  What  acids  does 
phosphorus  form?  What  proportion  of  the  air  is  nitrogen? 
Describe  a  method  for  preparing  chemical  nitrogen  as  dis- 
tinguished from  atmospheric  nitrogen.  What  is  formed  when 
an  ammonium  compound  is  heated?  What  might  be  put  with 
an  ammonium  compound  in  order  to  hold  the  acid  in  combina- 
tion and  allow  the  ammonia  to  escape?  Describe  the  tests  for 
both  free  and  combined  ammonia.  How  many  grams  of 
ammonium  chloride  would  be  necessary  to  form  ten  liters  of 
ammonia  ? 

Exercise  No.  20. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  How  many  liters  of  nitrous  oxide  could  be  prepared 
from  ten  grams  of  ammonium  nitrate?  How  could  you  dis- 
tinguish nitrous  oxide  from  oxygen?  How  many  grams  of 
copper  would  be  necessary  to  prepare  ten  liters  of  nitric  oxide  ? 
What  acid  is  theoretically  formed  by  the  addition  of  an  acid  to 
a  nitrite?  Describe  the  preparation  of  nitric  acid.  What  is 
a  good  test  for  nitric  acid  ? 


—79— 
Exercise  No.  21. 

1.  A  substance  melts  when  heated  in  the  sealed  tube,  colors 
the  flame  yellow,  gives  a  white  curdy  precipitate  with  silver 
nitrate  which  is  soluble  in  ammonium  hydroxide,  gives  a  color- 
less gas  that  fumes  in  air  when  treated  with  cone,  sulphuric 
acid ;  what  is  it  ? 

2.  A  substance  when  heated  in  a  sealed  tube  gives  off  red 
fumes,  the  residue  melts  and  solidifies  to  a  yellow  solid  when 
cold.     It  gives  a  reddish  color  to  the  flame ;    it  is  soluble  in 
water.      The   water    solution    gives    a   black    precipitate    with 
hydrogen  sulphide  which  does  not  dissolve  in  hydrochloric  acid, 
a  yellow  precipitate  with  potassium  chromate,  a  white  precip- 
itate   with    sulphuric    acid.      The    solid    treated    with    cone, 
sulphuric  acid  gives  off  a  colorless  liquid  that  distills  up  on  the 
side  of  the  test  tube.     A  piece  of  copper  placed  in  this  mixture 
causes  an  evolution  of  red  fumes.     What  is  the  substance? 

Exercise  No.  22. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  Give  the  names  of  all  the  elements  of  the  fifth  group. 
What  oxides  do  they  form?  Which  are  basic  and  which 
acidic?  What  acid  does  phosphorus  trioxide  form  with  water? 
Name  three  acids  that  phosphorus  pentoxide  forms  with 
water.  In  2,  why  does  the  phosphorus  take  fire  so  readily? 
What  is  formed  when  red  phosphorus  is  dissolved  in  nitric 
acid?  73.  How  many  liters  of  nitric  oxide  would  be  formed 
if  ten  grams  of  phosphorus  were  dissolved  in  nitric  acid? 
What  oxides  does  arsenic  form?  Are  these  oxides  basic  or 
acidic?  What  acids  would  these  oxides  form  with  water? 
Describe  two  different  forms  of  arsenic.  What  is  formed  when 
arsenic  burns  in  air?  What  is  formed  when  arsenic  is  dis- 
solved in  nitric  acid?  What  is  formed  when  arsenic  trioxide 
dissolves  in  alkalies?  What  acids  would  antimony  form? 

Exercise  No.  23. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  are  some  of  the  products  formed  when  wood 
and  paper  are  heated  in  a  sealed  tube?  What  classes  of  impuri- 


ties  does  charcoal  remove  from  solution?  How  many  liters  of 
carbon  dioxide  would  be  formed  if  ten  grams  of  lead  oxide 
were  reduced  to  lead  by  charcoal  ?  What  are  some  of  the  prod- 
ucts formed  when  soft  coal  is  heated  in  a  sealed  tube?  How 
is  coke  made  ?  How  is  charcoal  made  ?  Write  equations  show- 
ing the  reduction  of  metallic  oxides  with  carbon. 

Exercise  No.  24. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  oxides  does  carbon  form?  What  oxide  does 
carbon  form  when  burned  in  an  excess  of  oxygen?  How  many 
grams  of  carbon  would  be  necessary  to  form  ten  liters  of  carbon 
dioxide  at  normal  temperature  and  pressure?  State  three 
ways  in  which  carbon  dioxide  may  be  prepared.  What  acid 
does  this  oxide  form  with  water?  Is  it  a  stable  or  unstable 
acid?  What  are  the  black  specks  which  are  formed  when 
magnesium  burns  in  a  mixture  of  carbon  dioxide  and  air? 
What  is  formed  when  carbon  dioxide  is  passed  into  lime  water  ? 
What  are  the  two  products  which  would  be  formed  if  carbon 
dioxide  were  passed  in  excess  into  sodium  hydroxide  solution? 
What  is  formed  when  sodium  bicarbonate  is  heated?  Describe 
some  properties  of  calcium  chloride 

Exercise  No.  25. 

What  is  a  polybasic  acid  ;  a  dibasic  acid ;  a  tribasic  acid ;  an 
acid  salt ;  a  normal  salt,  and  a  basic  salt?  In  I,  what  proof  have 
you  that  the  product  was  hydrogen  sodium  carbonate?  What 
is  the  essential  composition  of  a  baking  powder?  Write  equa- 
tions showing  the  action  of  several  different  kinds  of  baking 
powders.  What  salt  is  the  product  of  the  action  of  a  cream  of 
tartar  powder  ? 

Exercise  No.  26. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  product  is  formed  when  sodium  is  heated  with 
silicon  dioxide?  What  is  the  action  of  silicon  on  sodium 
hydroxide?  WThat  acids  does  silicon  form?  How  could  you 
make  silicon  dioxide  from  sodium  silicate?  Is  silicon  dioxide 
a  basic  or  an  acidic  oxide?  State  three  ways  of  dissolving 


—Si- 
sand.     What  does   glass   consist  of?     What  is  the  action  of 
hydrofluoric  acid  on  sand?     How  many  liters  of  silicon  tetra- 
fluoride  would  be  formed  by  the  action  of  hydrofluoric  acid  on 
ten  grams  of  silicon  dioxide? 

Exercise  No.  27. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  are  the  elements  of  the  fourth  group?  What 
oxides  do  tin  and  lead  form  ?  Are  these  oxides  basic  or  acidic  ? 
What  acids  does  tin  form?  What  salt  is  formed  when  tin  is 
dissolved  in  hydrochloric  acid?  How  many  liters  of  hydrogen 
would  be  formed  if  ten  grams  of  tin  were  dissolved?  Explain 
the  changes  that  take  place  when  stannous  chloride  is  gradually 
added  to  a  solution  of  mercuric  chloride.  What  is  the  color 
of  stannous  sulphide?  When  this  is  dissolved  in  yellow 
ammonium  sulphide,  what  compound  is  formed?  When  this 
solution  is  acidified  with  hydrochloric  acid,  what  is  the  precip- 
itate? What  is  the  color  of  stannic  sulphide?  How  could 
you  change  stannous  sulphide  to  stannic  sulphide?  Explain 
the  effect  of  adding  a  piece  of  zinc  to  a  solution  of  stannous 
chloride.  121.  What  will  dissolve  lead?  In  what  acids  will 
it  not  dissolve?  Describe  lead  sulphide,  lead  sulphate,  lead 
chromate.  How  many  grams  of  zinc  would  precipitate  ten 
grams  of  lead  from  solution?  How  many  grams  of  lead  would 
produce  ten  liters  of  nitric  oxide? 

Exercise  No.  28. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  reactions  are  involved  in  the  Solvay  Process? 
How  would  you  form  calcium  and  barium  carbonates?  How 
do  sodium  and  potassium  color  the  Bunsen  flame?  Which 
color  would  obscure  the  other?  How  is  it  possible  to  see  the 
potassium  color  in  the  presence  of  sodium?  How  is  potassium 
nitrate  made?  How  many  grams  of  bicarbonate  of  soda  would 
be  necessary  to  give  ten  liters  of  carbon  dioxide  when  heated? 

Exercise  No.  29. 

Represent  by   equations,   all  the   reactions   involved   in   this 
exercise.      Name    the    alkaline    earth    metals.      Describe    the 
6 


—82— 

appearance  of  calcium  hydroxide.  How  is  lime  made?  How 
soluble  is  calcium  hydroxide?  What  is  lime  water  and  how  is 
it  made?  How  is  sodium  hydroxide  made?  How  did  you 
make  calcium  carbonate?  How  did  you  make  strontium  sul- 
phate? How  do  calcium  compounds  color  the  flame?  How- 
could  you  distinguish  between  a  calcium  and  a  strontium  com- 
pound? A  hydrochloric  acid  solution  contains  20%  acid  and 
has  a  specific  gravity  of  i.i,  how  many  c.c.  of  such  a  solution 
would  be  necessary  to  dissolve  ten  grams  of  calcium  hydrox- 
ide? 155. 

Exercise  No.  30. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  oxide  does  magnesium  form?  Is  it  basic  or 
acidic?  What  is  formed  when  it  dissolves  in  water?  How 
many  c.c.  of  a  hydrochloric  acid  solution  having  a  specific 
gravity  of  i.i  and  containing  20%  acid  would  be  necessary 
to  form  ten  liters  of  hydrogen  when  acting  on  a  metal?  155. 
What  precipitate  is  formed  wrhen  ammonium  hydroxide  is 
added  to  a  solution  of  magnesium  chloride?  147.  What  is 
the  nature  of  the  action  of  dilute  nitric  acid  on  zinc?  What 
is  the  precipitate  that  is  formed  at  first  when  sodium  hydrox- 
ide acts  on  a  solution  of  a  zinc  salt?  Has  this  precipitate  any 
acid  properties?  Why  does  it  dissolve  when  an  excess  of 
sodium  hydroxide  is  added?  What  compound  is  formed? 
How  could  you  separate  zinc  and  copper  by  means  of  hydrogen 
sulphide?  What  is  the  color  of  cadmium  sulphide?  What 
two  nitrates  does  mercury  form  and  under  what  circumstances 
would  each  result? 

Exercise  No.  31. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  is  the  nature  of  the  action  of  dilute  nitric  acid 
on  copper?  How  did  you  make  cupric  oxide?  State  three 
general  methods  for  the  preparation  of  metallic  oxides.  How 
did  you  make  cuprous  oxide?  Why  does  iron  precipitate  copper 
from  solution?  121.  What  is  the  most  striking  reaction  of 
soluble  copper  compounds?  What  is  the  nature  of  the  action 
of  nitric  acid  on  silver?  How  can  you  distinguish  between 


-83- 

silver  and  lead  chloride?  How  could  you  separate  silver,  lead 
and  mercurous  mercury?  How  many  c.c.  of  hydrochloric  acid 
solution  having  a  specific  gravity  of  i.i  and  containing  20% 
acid  would  be  necessary  to  precipitate  all  the  silver  in  100  c.c. 
of  a  silver  nitrate  solution  having  a  specific  gravity  of  i.oi  and 
containing  i%  silver  nitrate? 

Exercise  No.  32. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  oxides  do  iron  and  manganese  form?  Are 
these  basic  or  acidic?  Does  manganese  form  an  acidic  oxide? 
How  did  you  make  ferrous  sulphate?  How  else  might  you 
have  made  it?  What  is  green  vitriol?  Describe  ferrous 
hydroxide.  How  could  you  change  ferrous  sulphate  into  ferric 
sulphate?  How  might  you  change  ferrous  chloride  to  ferric 
chloride?  How  change  ferric  chloride  to  ferrous  chloride? 
What  precipitate  is  formed  when  hydrogen  sulphide  is  passed 
into  ferric  chloride  solution  ?  How  would  you  detect  the  pres- 
ence of  either  ferrous  or  ferric  iron,  each  in  the  presence  of 
the  other?  What  is  the  action  of  manganese  dioxide  on 
sulphurous  acid;  of  potassium  permanganate?  Give  the 
formulas  of  ten  oxidizing  agents  and  state  how  they  oxidize. 

Exercise  No.  33. 

Represent  by  equations,  all  the  reactions  involved  in  this 
exercise.  What  oxides  do  aluminum  and  chromium  form? 
What  reagents  will  dissolve  aluminum?  Is  aluminum  very 
basic?  Does  it  ever  act  acidic?  What  is  the  general  nature 
of  an  alum?  What  elements  form  alums?  How  do  aluminum 
salts  in  solution  act  toward  litmus  paper?  148.  How  could 
you  change  basic  chromium  to  acid  chromium?  Describe  the 
changes  that  take  place  when  alcohol,  potassium  dichromate 
and  sulphuric  acid  are  heated  together?  How  many  c.c.  of  a 
potassium  dichromate  solution  having  a  specific  gravity  of  1.03 
and  containing  5%  of  the  salt  would  be  necessary  to  oxidize 
all  the  iron  in  100  c.c.  of  a  ferrous  sulphate  solution  having  a 
specific  gravity  of  1.05  and  containing  10%  of  green  vitriol? 


-84- 

Exercise  No.  34. 

What  metals  color  the  flame  red:  yellow;  blue  or  green; 
purple?  What  are  the  insoluble  chlorides?  What  are  the 
insoluble  sulphates?  What  insoluble  sulphides  are  white; 
brown;  black;  yellow;  orange;  pink?  What  oxide  is  yellow 
when  hot  and  white  when  cold?  What  insoluble  hydroxides 
are  white;  bluish;  brown;  green?  What  substance  in  solu- 
tion gives  a  precipitate  of  sulphur  as  well  as  an  odor  of  sulphur 
dioxide  when  treated  with  an  acid  ? 

Exercise  No.  35. 

What  are  the  general  methods  for  making  basic  oxides? 
State  seven  methods  for  making  salts.  Before  selecting  a 
method  for  making  a  salt,  what  particular  things  must  be  con- 
sidered? How  are  insoluble  salts  made?  How  are  the  soluble 
hydroxides  made?  How  are  the  insoluble  hydroxides  made? 
How  could  an  acid  salt  be  made  ? 


.  PART  II. 
FUNDAMENTAL    IDEAS, 


DEFINITIONS. 

1.  Matter  is  that  which  has  weight. 

2.  An  element  is  a  substance  that  contains  only  one  kind  of 
matter.     There  are  about  seventy-five  elements,  of  which  only 
about  twenty-five  are  common. 

Examples  of  common  elements  are:    iron,  copper,  lead,  tin, 
gold,  and  silver.     167-168. 

3.  A  symbol  of  an  element  is  a  sign  that  stands  for  that 
substance.     It  ordinarily  consists  of  one  or  two  characteristic 
letters  of  the  English  or  Latin  name. 

4.  Symbols  and  names  of  some  common  elements,  in  groups. 
167. 

GROUP  i.  GROUP  2.  GROUP  3.        GROUP  4. 

H,  Hydrogen.     Mg,  Magnesium.    B,  Boron.        C,  Carbon. 
Na,  Sodium.  Ca,  Calcium.        Al,  Aluminum.  Si,  Silicon. 

K,  Potassium.      Zn,  Zinc.  Sn,  Tin. 

Ammonium.  Ba,  Barium.  Pb,  Lead. 


Ag,  Silver. 

GROUP  5. 

N,  Nitrogen. 

P,  Phosphorus. 
As,  Arsenic. 
Sb,  Antimony. 
Bi,  Bismuth. 


Hg,  Mercury. 

GROUP  6.  GROUP  7.  GROUP  8. 

O,  Oxygen.  F,  Fluorine.  Fe,  Iron. 

S,  Sulphur.          Cl,  Chlorine.  Ni,  Nickel. 

Cr,  Chromium.     Br,  Bromine.  Co,  Cobalt. 
I,  Iodine. 


*  Not  an  element,  but  a  compound  of  N  and  H  that  acts  like  one. 


—86— 

5.  Properties  of  matter.    Those  qualities  which  are  peculiar 
to  it  and  characteristic  of  it.     An  exact  description  of  any  kind 
of  matter  involves  telling  what  it  looks  like,  smells  like,  tastes 
like,  how  it  behaves  at  different  temperatures  and  when  mixed 
with  other  substances. 

6.  All  matter  is  supposed  to  be  made  up  of  a  very  large 
number  of  very  small  particles  called  molecules,  and  these  mole- 
cules to  be  made  up  of  still  smaller  particles  called  atoms. 

7.  A  molecule  is  the  smallest  particle  of  matter  that  can 
exist  and  still  retain  all  the  properties  of  that  kind  of  matter. 
If  the  substance  is  an  element,  the  molecule  will  consist  of  one 
or  more  atoms  of  that  element;   if  the  substance  is  a  chemical 
compound,  the  molecule  will  contain  at  least  one  atom  of  every 
element  contained  in  the  substance. 

8.  An  atom  is  the  smallest  particle  of  an  element  which 
forms  part  of  a  molecule. 

9.  The  chemical  composition  of  a  substance  means  the 
relative  proportions,  by  weight,  of  all  the  elements  contained 
in  it. 

10.  A  physical  change  is  one  in  which  the  chemical  com- 
position of  the  body  is  not  altered.     Thus,  a  piece  of  matter  may 
be  set  in  motion,  be  heated,  electrified,  magnetized,  melted  or 
dissolved  without  in  any  way  changing  its  chemical  composition. 
Such  changes  are  physical. 

11.  A  chemical  change  is  one  in  which  the  chemical  com- 
position of   the  body   is   altered.      Such   changes    are   always 
.accompanied   by   some   change   in   the   total   energy   which   is 
associated  with  the  reacting  substances.     That  is,  there  is  an 
evolution   or    absorption   of    heat,    light   or    electricity.      For 
example ;  if  sugar  be  heated  hot  enough,  it  changes  into  carbon 
and  water.     Or  if  carbon  be  heated  in  the  air,  it  changes  to  a 
colorless  gas,  called  carbon  dioxide.     The  rusting  of  iron,  the 
souring  of  milk,  and  the  fermenting  of  cider  are  also  examples 
of  chemical  changes. 

12.  If  two  or  more  elements  be  mixed  in  such  a  way  that 
they  cannot  be  separated  again  by  mechanical  means,  a  chemical 
compound  is  said  to  be  formed.     For  example ;  if  sulphur  and 
iron  be  heated  together,   a  compound  called  iron  sulphide  is 
formed;    and  no  mechanical  process  can  again  separate  them. 
Such  a  compound  contains  only  one  kind  of  molecule. 


-87- 

13.  A  mechanical  mixture  is  one  containing-  the  elements 
mixed  in  such  a  way  that  they  may  be  separated  by  mechanical 
means.     The  iron  and  sulphur  could  readily  have  been  separated 
by  means  of  a  magnet  before  they  had  been  heated.     Most 
mechanical    mixtures    are    mixtures    of    chemical    compounds. 
When  viewed  under. the  microscope  such  mixtures  show  dif- 
ferent kinds  of  particles.     In  the   case  of  solutions  there  is 
considerable  difficulty  in  distinguishing  them  from  chemical  com- 
pounds, for  all  parts  of  the  same  solution  have  exactly  the  same 
chemical  composition  and  it  is  impossible  to  distinguish  differ- 
ent kinds  of  articles  by  means  of  a  microscope.     However,  the 
solvent  can  usually  be  entirely  evaporated  in  the  pure  condition, 
leaving  the  other  material  by  itself,  either  in  the  liquid  or  solid 
condition.     A  mechanical  mixture  contains  two  or  more  differ- 
ent kinds  of  molecules. 

14.  An  atomic  weight  is  a  number  expressing  the  ratio 
between  the  weight  of  the  smallest  particle  of  an  element  form- 
ing part  of  a  molecule  and  the  weight  of  an  atom  of  hydrogen. 
In  other  word's,  it  is  the  ratio  of  the  amount  of  matter  in  an  atom 
to  the  amount  of  matter  in  one  atom  of  hydrogen.     Sometimes 
one-sixteenth  the  weight  of  the  oxygen  atom  is  taken  as  the 
anit,  in  place  of  the  hydrogen  atom. 

15.  A  molecular  weight  is  a  number  which  shows  how 
many  times  heavier  a  molecule  of  a  substance  is  than  one  atom 
of  hydrogen.     It  is  equal  to  the  sum  of  the  atomic  weights  of 
all  the  atoms  in  the  molecule. 


FULL  MEANING  OF  SYMBOLS  AND 
FORMULAS. 

1 6.  Full  meaning  of  the  symbol  of  an  element. 

1.  It  is  a  sign  that  stands  for  that  substance. 

2.  It  is  a  sign  that  stands  for  one  atom  of  that  substance. 

3.  It  is  a  sign  that  stands  for  a  certain  weight  of  that  sub- 
stance. 

For  example:  S  is  a  sign  that  stands  for  sulphur,  for  one 
atom  of  sulphur,  and  for  32  weights  of  sulphur. 

The  weight  corresponding  to  a  given  symbol  is  the  same  as 
the  atomic  weight  of  the  element,  and  may  be  found  in  any  table 
of  elements.  168. 

17.  The  formula  of  a  compound  is  a  sign  that  stands  for 
one  molecule  of  that  substance.     It  is  made  up  of  the  symbols 
of  the  elements  contained  in  the  compound,   with   a  number 
below  and  at  the  right  of  each  symbol,  indicating  how  many 
atoms  of  that  element  are  contained  in  the  molecule. 

1 8.  Formulas  of  some  common  compounds. 
H2O,  Hydrogen  oxide  or  water. 

CO2,  Carbon  dioxide  or  carbonic  acid  gas. 

MgO,  Magnesium  oxide  or  magnesia. 

NaCl,  Sodium  chloride  or  common  salt. 

HC1,  Hydrogen  chloride,  hydrochloric  acid  or  muriatic  acid. 

HNO3,  Hydrogen  nitrate  or  nitric  acid. 

H2SO4,  Hydrogen  sulphate,  sulphuric  acid  or  oil  of  vitriol. 

The  Full  Meaning  of  the  Formula  of  a  Compound. 

19.  This  can  best  be  shown  by  giving  the  exact  meaning  of 
some  formula. 

1.  H2SO4  is  a  sign  that  stands  for  sulphuric  acid. 

2.  It  stands  for  one  molecule  of  sulphuric  acid. 

3.  It  stands  for  98  weights  of  sulphuric  acid. 
It  states  that,  • 

4.  Sulphuric  acid  is  made  of  hydrogen,  sulphur  and  oxygen. 

5.  That  one  molecule  of  sulphuric  acid  contains  two  atoms 
of  hydrogen,  one  of  sulphur  and  four  of  oxygen.- 


—89— 

6.  That  these  elements  are  present  in  the  proportion  of  2 
weights  of  hydrogen,  32  weights  of  sulphur  and  64  weights  of 
oxygen. 

The  student  should  practice  giving  the  full  meaning  of  other 
formulas. 

20.  Analysis  is  the  process  of  rinding  out  the  relative  pro- 
portions, by  weight,  of  the  elements  contained  in  a  substance. 

The  results  of  an  analysis  are  always  expressed  in  per  cents. 

21.  To  change  the  Composition  of  a  Substance  as  shown 
by  a  Formula  to  a  percentage  Basis. 

Example.  Given  the  formula  of  sulphuric  acid,  H2SO4,  to 
find  its  percentage  composition. 

2H=    2 


98 
There  are  98  weights  in  the  molecule  altogether;  of  these  2  are 

2 

hydrogen;  therefore  —  reduced  to  a  decimal  and  multiplied  by 
98 

100  must  equal  the  percentage  of  this  constituent. 

Proceeding  in  the  same  way  with  the  other  elements  we  have, 

H=  "8=  .0204  or    2.04$ 
S=j*|=.3265  or  32.65$ 


O=-*=.653i  or  65.31$ 

100.  OO 

Exercise. 

22.  Compute  the  percentage  composition  of  substances  from 
their  formulas. 

Derivation  of  simple  Formulas  from  Analysis. 

23.  If  the  molecule  happens  to  contain  a  single  atom  of  any 
one  kind,  it  is  a  simple  matter  to  derive  the  formula  from  the 
aiaalysis.     It  is  only  necessary  to  divide  each  given  percentage 


by  the  corresponding  atomic  weight  and  the  results  will  be  the 
relative  numbers  of  atoms  in  the  molecule.  Each  number  will 
be  a  simple  multiple  of  the  smallest,  and  the  actual  numbers  of 
atoms  is  readily  seen  by  inspection.  For  the  treatment  of  cases 
involving  more  complex  molecules,  see  95. 

Exercise. 

24.  Derive   the   formulas   corresponding   to   the    following 
analyses : 

Hydrogen,  2.74%,  chlorine,  97.26%;  hydrogen,  11.11%, 
oxygen,  88.89%;  hydrogen,  17.63%,  nitrogen,  82.35%; 
hydrogen,  25.00%,  carbon,  75.00%. 

25.  Metals  and  Non-metals.     168. 

All  the  elements  may  be  divided  into  two  classes;  metals 
and  non-metals.  In  addition  to  the  common  elements  like 
copper,  lead,  tin,  iron,  and  silver,  the  student  may  remember 
that  mercury,  bismuth  and  antimony  are  metals.  Also  that 
most  of  the  elements  whose  names  end  in  "um"  are  metals. 
168. 

26.  A    binary    compound    is    one    containing    only    two 
elements. 

In  naming  a  binary  compound  the  name  of  the  metal  is 
placed  first,  unchanged,  followed  by  the  name  of  the  non-metal 
with  its  name  changed  to  an  ending  in  "ide,"  e.g.,  a  compound 
of  sodium  and  chlorine  is  called  sodium  chloride;  or  of  iron 
and  oxygen,  iron  oxide,  etc.  While  almost  any  element  may 
form  a  binary  compound  with  any  other  element,  by  far  the 
most  common  compounds  are  those  of  the  metals  with  the 
non-metals. 

Where  both  of  the  elements  are  non-metals,  the  name  of 
either  may  be  placed  first,  followed  by  the  name  of  the  other 
with  an  ending  in  "ide,"  e.g.,  hydrogen  nitride  or  nitrogen 
hydride. 

27.  The  Law  of  Definite  Proportions. 

If  equal  weights  of  two  elements  be  caused  to  combine  it  will 
be  found  that  one  of  them  will  entirely  enter  into  combination 
and  some  of  the  other  will  be  left  uncombined.  No  matter  how 


the  quantities  be  varied  one  of  the  substances  will  always 
entirely  enter  into  combination.  If  the  resulting  compound  be 
analysed,  the  constituent  elements  will  always  be  found  to  be 
present  in  definite  proportions  by  weight.  If  the  two  original 
elements  be  combined  in  these  proportions  they  will  exactly 
unite  to  form  the  compound.  Thus  if  four  weights  of  sulphur 
and  seven  of  iron  be  heated  together  they  exactly  combine  to 
form  a  compound  known  as  iron  sulphide. 

In  a  given  chemical  compound  the  elements  are  always 
present  in  the  same  fixed  proportions  by  weight. 

28.  The  Law  of  Multiple  Proportions. 

Analysis  of  all  the  known  binary  compounds  shows  that  there 
are  many  cases  of  the  same  two  elements  uniting  in  different 
proportions.  In  such  cases  it  is  found  that  when  a  fixed 
quantity  of  one  element  unites  with  variable  quantities  of 
another,  the  weights  of  the  variable  element  are  simple  multi- 
ples of  the  smallest.  Thus,  seven  weights  of  nitrogen  combine 
with  four,  eight,  twelve  and  sixteen  weights  of  oxygen,  forming 
four  different  oxides. 

29.  The  Atomic  Hypothesis  of  Dalton. 

As  early  as  the  fifth  century  B.  C.  certain  philosophers  con- 
sidered all  matter  to  be  made  up  of  small  particles  called  atoms. 
These  atoms  were  all  assumed  to  be  of  the  same  kind  and  the 
same  size.  Different  kinds  of  matter  were  supposed  to  result 
from  the  atoms  approaching  or  receding  from  each  other. 

In  1807  Dalton  proposed  the  atomic  theory  which  is  in  use 
today  and  is  the  basis  of  all  modern  theoretical  chemistry. 
According  to  his  theory,  all  the  atoms  of  a  given  element  are  of 
the  same  size  and  the  same  weight  but  atoms  of  different  ele- 
ments have  different  weights.  If  two  or  more  atoms  of  the 
same  kind  unite,  they  constitute  a  molecule  of  an  element.  If 
two  or  more  atoms  of  different  kinds  unite,  they  form  a  mole- 
cule of  a  compound. 

This  theory  agrees  with  the  law  of  definite  proportions  and 
the  law  of  multiple  proportions,  for  if  two  atoms  unite  to  form 
a  molecule  of  a  compound,  the  proportions  of  the  elements  in  a 
mass  of  the  substance  would  be  identical  with  the  proportions 


--92— 

by  weight  of  the  elements  in  the  molecule.  Again,  if  a  fixed 
weight  of  one  element  unites  with  variable  weights  of  another 
element  to  form  several  different  compounds,  the  ratios  of  the 
weights  of  the  variable  element  to  each  other  are  of  the  nature 
of  simple  numbers. 

This  is  well  illustrated  by  the  oxides  of  nitrogen.  28  weights 
of  nitrogen  combine  with  16,  32,  48,  64  and  80  weights  of 
oxygen.  The  ratio  of  the  last  four  weights  to  the  first  is  in 
every  case  a  simple  whole  number. 


COMBINING  POWER  OR  VALENCE. 

30.  It  is  possible  to  determine  exactly  the  molecular  weight 
of  a  compound  by  several  processes.     92-93.     This,  taken  in 
connection  with  the  analysis  and  the  atomic  weights,  makes  it 
possible  to  arrive  at  the  correct  formula.     95. 

The  formulas  of  all  the  binary  compounds  have  been 
determined;  and  from  an  inspection  of  them  it  appears,  that 
the  atoms  of  the  various  elements  have  different  powers  of 
holding  each  other  in  combination  in  the  molecules. 

31.  Combining  power  or  Valence  is  the  power  that  an 
atom  has  of  combining  with  other  atoms  to  form  molecules. 
All  the  elements  in  the  same  group  have  the  same  combining 
power.     The  valence  of  an  element  is  measured  by  the  number 
of  atoms  of  hydrogen,  or  their  equivalent,  which  its  atom  can 
hold  in  combination.     Compounds  with  oxygen  or  hydrogen  are 
usually  taken  to  show  the  valence,  although  any  element  whose 
valence  is  known  will  serve  just  as  well. 

Combining  power  of  the  elements  by  groups. 

The  elements  in  group  one  have  a  power  of  one. 

The  elements  in  group  two  have  a  power  of  two. 

The  elements  in  group  three  have  a  power  of  three. 

The  elements  in  group  four  have  a  power  of  four. 

The  elements  in  group  five  have  powers  of  three  and  five. 

The  elements  in  group  six  have  powers  of  two,  four  and 
six,  except  oxygen  which  has  only  two. 

The  elements  in  group  seven  have  powers  of  one,  three, 
five  and  seven. 

The  elements  in  group  eight  (see  167)  have  powers  of  two 
and  three. 

In  cases  where  an  element  has  several  combining  powers, 
the  one  to  be  used  in  the  following  exercises,  unless  stated  to  the 
contrary,  is  the  lowest. 

DERIVATION  OF  FORMULAS  OF  BINARY  COM- 
POUNDS FROM  THE  KNOWN  COMBINING 
POWERS  OF  THE  ELEMENTS. 

32.  Almost  any  two  elements  can  combine  to  form  a 
binary  compound.      The  total  valence  or  combining  power 
of  the  atoms  of  any  one  kind  forming  a  molecule  of  a  binary 


—94— 

compound  is  the  least  common  multiple  of  the  two  com- 
bining powers  concerned. 

In  other  words,  there  must  be  enough  atoms  of  the  first  kind 
present  to  give  a  total  valence  equal  to  that  of  one  or  more 
atoms  of  the  second  kind.  If  this  valence  is  the  least  possible 
one  that  gives  equivalent  combining  powers,  the  resulting  for- 
mula is  the  simplest  one  possible. 

The  simplest  formula  is  not  necessarily  the  correct  one. 
The  true  formula  may  be  a  multiple  of  the  simplest.  95. 

Sodium  has  a  power  of  one,  being  in  the  first  group,  and  it 
will  therefore  exactly  combine  with  one  atom  of  chlorine,  which 
is  in  the  seventh  group,  but  has  a  low  common  power  of  one. 
The  formula  would  be  NaCl. 

Magnesium,  with  a  power  of  two,  will  combine  with  two 
chlorine  atoms,  giving  the  formula  MgCl2. 

Aluminum,  in  the  third  group,  with  a  power  of  three,  would 
combine  with  three  chlorine  atoms,  giving  the  formula  A1C13. 

The  formula  of  a  compound  of  carbon  and  chlorine  would  be 
CC14. 

The  formula  of  a  compound  of  aluminum  and  oxygen  would 
be  A12O3,  for  the  least  common  multiple  of  the  two  combining 
powers  would  be  six,  and  it  takes  two  aluminum  atoms  to  give 
an  equivalent  power.  Oxygen,  in  the  sixth  group,  always  has 
a  combining  power  of  two  and  it  will  take  three  atoms  to  give 
a  valence  of  six. 

The  sub-numbers  in  the  formula  are  ordinarily  the  same 
as  the  combining  powers  of  the  two  elements,  reversed.  If 
one  is  divisible  by  the  other,  the  simplest  ratio  is  taken  in  the 
reversed  order,  e.g.,  carbon  has  a  power  of  four  and  oxygen  of 
two;  the  formula  is  not  C2O4,  but  CO2. 

When  two  elements  form  several  different  binary  com-. 
pounds,  it  is  customary  to  distinguish  between  them  by  placing 
the  prefixes,  mon- ;  di-  ;  tri-  ;  tetra- ;  penta- ;  etc.,  before  the 
name  of  the  variable  element.  Thus  SO2  is  sulphur  dioxide, 
SO3  is  sulphur  trioxide,  P2O5  is  phosphorus  pentoxide,  P2O3 
is  phosphorus  trioxide. 

Remember  always,  that  formulas  are  not  originally  derived 
from  valence  but  valence  from  formulas.  And  that  the  theory 
of  valence  is  only  taught  in  order  to  relieve  the  student  of  the 


—95— 

necessity  of  remembering  the  combining  proportions  of  the  ele- 
ments in  thousands  of  well  known  compounds. 

Exercise. 

33.     Give  the  formulas  and  names  of  all  the  binary  com- 
pounds of  the  metals  with  the  non-metals. 


CHEMICAL  EQUIVALENCE. 

34.  A  quantity  of  one  element  is  said  to  be  equivalent  to  a 
certain  quantity  of  another  element  when  there  are  enough 
atoms  of  each  present  to  give  the  same  combining  power. 

One  atom  of  sodium  is  equivalent  to  one  atom  of  potassium. 

Two  atoms  of  sodium  are  equivalent  to  one  atom  of  mag- 
nesium. 

Two  atoms  of  magnesium  are  equivalent  to  one  atom  of 
carbon. 

Three  atoms  of  zinc  are  equivalent  to  two  atoms  of  aluminum. 

The  term  "atom"  is  used  in  the  sense  of  a  certain  number  of 
weights.  Thus,  three  atoms  of  zinc  means  three  times  65 
weights  of  zinc ;  and  two  atoms  of  aluminum  means  two  times 
27  weights  of  aluminum. 

The  Number  of  Grams  of  any  one  Element  equivalent  to 
a  Certain  Number  of  Grams  of  any  other  Element. 

How  many  grams  of  aluminum  are  equivalent  to  ten  grams 
of  zinc? 

Two  atoms  of  aluminum  are  equivalent  to  three  atoms  of 
zinc.  That  is,  54  weights  of  aluminum  are  equivalent  to  195 
weights  of  zinc.  One  weight  would  therefore  be  equivalent  to 

—  as  much,  or  — -,  and  as  many  would  be  equivalent  to  ten 
54  54 

as  -^  is  contained  in  ten,  or  loX — --. 
54  195 

Or,  since  195  weights  of  zinc   are  equivalent  to  54  weights 

of  aluminum,  one  weight  would  be  equivalent  to  -  —  as  much 
or ><54,  and  ten  weights  would  be  equivalent  to  zoX  — - . 


— g<5— 

Exercise. 

35.  Give    the    chemical    equivalence    between    every    two 
elements,   and  give  the  expression   for  the  answer  when  ten 
grams  of  either  element  is  given.     That  is,  ten  grams  of  sodium 
is  equivalent  to  how  many  grams  of  aluminum,  etc.? 

36.  A  reaction  is  a  chemical  change. 

37.  A  reagent  is  any  substance  that  takes  part  in  a  reaction. 

38.  A  chemical  equation  is  one,  on  the  left  hand  side  of 
which  are  placed  the  symbols  of  the  molecules  of  the  substances 
used,  and  on  the  right  hand  side  are  placed  the  symbols  of  the 
molecules  of  the  substances  formed.      A  number  is  placed  before 
each  symbol  indicating  the  least  possible  number  of  molecules  of 
that  substance  which  could  take  part  in  that  reaction. 

An  equation  can  never  be  a  complete  description  of  a  reaction 
for  it  only  tells  what  substances  take  part  in  the  reaction,  and 
their  proportions.  Besides  these  facts,  a  complete  description 
must  state  the  circumstances  under  which  the  particular  reac- 
tion under  consideration  takes  place,  especially  with  regard  to 
the  temperature,  the  pressure  and  the  concentrations  of  the 
various  substances  concerned. 

According  to  this  definition,  it  is  necessary  to  know  how 
many  atoms  there  are  in  the  molecules  of  the  elements.  As 
there  is  considerable  uncertainty  as  to  the  number  of  atoms  in 
the  molecules  of  many  of  the  elements,  for  the  present  it  will 
be  stated  that  the  elementary  gases,  oxygen,  hydrogen,  nitro- 
gen, chlorine  and  several  other  elements  (see  168)  have  two 
atoms  in  the  molecule,  and  when  these  elements  appear  by 
themselves  in  an  equation  they  should  be  written,  O2,  H2,  N2 
and  C12.  On  account  of  the  uncertainty  it  is  preferred  to  write 
all  other  elements  as  atoms. 

When  all  the  facts  regarding  a  chemical  reaction  are  known 
it  is  possible  to  represent  that  reaction  by  a  chemical  equation. 

For  example,  46  weights  of  sodium  exactly  unite  with  16 
weights  of  oxygen  to  form  62  weights  of  sodium  oxide.  Since 
the  atomic  weight  of  sodium  is  23,  it  will  take  two  atoms  of 
sodium  and  one  atom  of  oxygen  to  form  one  molecule  of  sodium 
oxide. 

That  could  be  expressed: 


—97— 

-f-  O  =  Na2O ;  but  since  we  are  going  to  express  oxygen 
in  the  form  of  a  molecule  we  shall  double  the  quantities  and 
write  it,  4Na  +  O2  =  2Na2O. 

Again,  to  be  more  general,  suppose  that  actual  experiment 
shows  that  26.47  grams  of  aluminum  exactly  combine  with 
23.53  grams  of  oxygen  to  form  50.00  grams  of  aluminum 
oxide,  and  that  it  is  desired  to  express  this  reaction  by  an  equa- 
tion. Dividing  the  first  two  numbers  by  the  corresponding 
atomic  weights  of  the  elements  and  the  last  by  the  molecular 
weight  of  aluminum  oxide,  we  obtain  three  numbers  which 
show  at  once  the  relative  numbers  of  atoms  of  the  elements 
compared  to  the  number  of  molecules  of  the  compound.  These 
numbers  are, 

Al  O  A12O3 

.9804          1.4700          .4902 

This  means  that  two  atoms  of  aluminum  unite  with  three 
of  oxygen  to  form  one  molecule  of  aluminum  oxide. 

Since  we  desire  to  write  a  molecular  equation  as  far  as  pos- 
sible we  double  all  these  quantities  and  write  the  equation, 
4A1  +  302  =  2A1203. 


EQUATIONS    REPRESENTING    THE    FORMATION 
OF   BINARY    COMPOUNDS. 

39.  Since  the  combining  power  of  the  various  elements  has 
been   well   established   by   experiment   in   the   past,   it   is   now 
possible  to  predict  their  formation  exactly  and  represent  the 
corresponding  reactions  by  equations. 

Nitrogen  unites  with  oxygen.  Two  nitrogen  atoms  are  chem- 
ically equivalent  to  three  oxygen  atoms,  and  unite  to  form  one 
molecule  of  nitrogen  oxide.  Thus: 

2N  -f-  30  =  N2O3,  but  in  this  equation  we  have  not  repre- 
sented oxygen  and  nitrogen  as  molecules ;  we  therefore  rewrite 
the  equation,  doubling  the  quantities ;  so, 
2N2  +  302  =  2N203. 

Exercise. 

40.  Write  equations  showing  the  formation  of  the  binary 
compounds  resulting  from  the  combinations  of  the  metals  with 

7 


-98- 

the  non-metals.     Especially  show  the  formation  of  the  oxides, 
sulphides,  and  chlorides. 

2Na  +  C12  =  2NaCl. 
4A1  +  302  =  2A1203. 
C  +  02  =  C02. 
Zn  +  S  =  ZnS. 
Etc. 

FULL  MEANING  OF  A  CHEMICAL  EQUATION. 

41.  The  equation, 

4A1  -f-  3O2  —  2A12O3  means 

1.  That  aluminum  combines  with  oxygen  to  form  aluminum 
oxide. 

2.  That  four  atoms  of  aluminum  exactly  unite  with  three 
molecules  of  oxygen  to  form  two  molecules  of  aluminum  oxide. 

3.  In  other  words,   108  weights  of  aluminum  exactly  unite 
with  96  weights  of  oxygen  to  form  204  weights  of  aluminum 
oxide. 

4.  Therefore  that   ten  grams   of  aluminum  would  exactly 

06  204. 

unite  with  — 0X  10  grams  of  oxygen  and  form  — 0X  10  grams 

lOo  lOo 

of  aluminum  oxide. 

Also  that  10  grams  of  oxygen  would  combine  with  ~r~X  10 

grams   of   aluminum   and  form  —' ^X  10  grams  of  aluminum 
oxide  and  finally  that  10  grams  of  aluminum  oxide  would  be 

formed  by  the  combination  of X  10   grams   of   aluminum 

204 

with  -•— Xio  grams  of  oxygen. 

42.  Give    the    full    meaning   of    every    equation   you    have 
written,  working  it  out  for  ten  grams  as  above. 

FORMULAS  OF  COMPOUNDS  CONTAINING  MORE 
THAN  TWO  ELEMENTS. 

43.  A  compound  containing  more  than  two  elements  is  called 
a  Ternary  compound. 

44.  It  is   found  that  when   the   electric   current   is   passed 
through   many    fused   compounds   or   compounds    in   solution, 


that  they  are  broken  up  into  two  parts ;  one  part,  consisting  of 
the  metal  or  hydrogen,  going  to  the  negative  electrode,  and  all 
the  rest  of  the  substance  going  to  the  positive  electrode.  124. 

45.  A  radical  is  an  atom  or  a  group  of  atoms  which  enters 
various  reactions  as  a  whole.     Ex. — H ;   Na ;   NH4 ;   Cl ;   SO4 ; 
NO3. 

46.  A   basic   radical   is    one   that   will   move   toward    the 
negative  electrode,  when  an  electric  current  is  passed  through 
a   solution  of  the   substance.      Hydrogen   and  nearly   all   the 
metals  constitute  the  basic  radicals. 

47.  An  acid  radical  is  one  that  moves  toward  the  posi- 
tive electrode,  when  an  electric  current  is  passed  through  a 
solution  of  the   substance.      Nearly   all   of   the   acid   radicals 
contain  several  atoms. 

48.  Radicals  may  be  considered  to  have  combining  powers 
exactly  like  the  elements  themselves,  and  to  unite  with  each 
Cither  to  form  compounds  in  a  similar  way. 

49.  Names  of  radicals.     The  names  of  the  radicals  given 
in  the  following  list  are  the  names  which,  put  together  just 
as  they  are,  will  give  the  proper  chemical  name  of  the  com- 
pound which  any  two  radicals  form. 

50.  In  order  that  the  student  may  become  familiar  at  once 
with  many  of  the  names  and  formulas  of  compounds  that  he 
must  necessarily  make  use  of  in  the  laboratory  work,  the  follow- 
ing: list  of  radicals  should  be  committed  to  memory.     Exten- 
sive drill  should  also  be  given  in  combining  the  basic  with  the 
acid  radicals  and  in  naming  the  compounds  so  indicated.     The 
derivation  of  these  and  other  radicals  will  be  given  later. 

LIST  OF  RADICALS.  WITH  NAMES. 
BASIC  RADICALS.  ACID  RADICALS. 

GROUP  i.     POWER  OF  ONE.         GROUP  i.     POWER  OF  ONE. 

H,  Hydrogen.  C2H3O2,  Acetate. 

Na,  Sodium.  Br,  Bromide. 

K,  Potassium.  C1O3,  Chlorate. 

NH4,  Ammonium.  Cl,  Chloride. 

Ag,  Silver.  OH,  Hydroxide. 

Hg,  Mercurous.  I,  Iodide. 

Cu,  Cuprous.  NO3,  Nitrate. 


— 100 

BASIC  RADICALS.  ACID  RADICALS. 

GROUP  2.    POWER  OF  Two.        GROUP  2.     POWER  OF  Two. 

Mg,  Magnesium.  CO3,  Carbonate. 

Ca,  Calcium.  O,  Oxide. 

Zn,  Zinc.  SO4,  Sulphate. 

Ba,  Barium.  S,  Sulphide. 

Hg,  Mercuric. 

Cu,  Cupric. 

Fe,  Ferrous. 

Pb,  Lead. 

GROUP  3.     POWER  OF  THREE.     GROUP  3.     POWER  OF  THREE. 

Al,  Aluminum.  PO4,  Ortho-phosphate. 

Fe,  Ferric.  Fe(CN)6,  Ferri cyanide. 

GROUP  4.     POWER  OF  FOUR. 
Fe(CN)6,  Ferrocyanide. 

51.  If  a  radical  has  two  different  combining  powers,  it  is 
evident  that  it  will  form  two  classes  of  compounds ;  and 
therefore  it  is  necessary  that  two  different  names  be  assigned 
to  it,  in  order  to  signify  which  compound  is  being  considered. 
Thus  iron  forms  two  oxides ;  ferrous  oxide,  with  the  formula 
FeO,  and  ferric  oxide,  with  the  formula  Fe2O3. 


COMBINATIONS    OF   RADICALS. 

52.     Basic  radicals  unite  with  acid  radicals. 

The  total  combining  power  of  the  basic  radical  must  be 
exactly  equal  to  the  total  combining  power  of  the  acid  radical, 
exactly  as  in  the  formation  of  a  binary  compound. 

When  it  is  necessary  to  represent  a  compound  radical  as 
occurring  several  times  in  a  formula,  the  whole  radical  is  put  in 
parenthesis  with  a  number  below  and  at  the  right,  showing  how 
many  times  it  is  found. 

The  name  is  formed  by  simply  putting  together  the  two 
names  of  the  radicals,  placing  that  of  the  basic  radical  first. 


— 101 

Exercise. 

53.  Give  the  names  and  formulas  of  all  the  radicals. 
Give  the  names  and  formulas  of  the  radicals  by  groups. 

Exercise. 

54.  Give  the  formulas  and  names  of  all  the  compounds  of 
the  basic  radicals  with  the  acid  radicals. 

To  illustrate,  take  one  typical  radical  from  each  group. 
H/  Cl/ 

Na.'  NO3/ 

Ba."  SO4." 

Al/"  PO4/" 

Fe(CN)6.""      • 

The  primes  indicate  the  combining  power  to  be  used  in  this 
particular  case. 

HC1,  Hydrogen  chloride. 
HNO3,  Hydrogen  nitrate. 
H2SO4,  Hydrogen  sulphate. 
H3PO4,  Hydrogen  phosphate. 
H4Fe(CN)6,  Hydrogen  ferrocyanide. 

In  this  set  it  takes  a  number  of  H's  equal  to  the  combining 
power  of  the  acid  radical  to  form  the  molecule. 

NaCl,  Sodium  chloride. 
NaNO3,  Sodium  nitrate. 
Na2SO4,  Sodium  sulphate. 
Na3PO4,  Sodium  phosphate. 
Na4Fe(CN)6,  Sodium  ferrocyanide. 

Here  again,  since  the  combining  power  of  sodium  is  one,  it 
takes  a  number  of  sodium  radicals  equal  to  the  combining 
power  of  the  acid  radical  to  form  the  molecule. 

BaCl2,  Barium  chloride. 
Ba(NO3)2,  Barium  nitrate. 
BaSO4,  Barium  sulphate. 
Ba3(PO4)2,  Barium  phosphate. 
Ba2Fe(CN)6,  Barium  ferrocyanide. 

Barium  has  a  power  of  two,  chloride  has  a  power  of  one; 
two  chloride  radicals  will  be  equivalent  to  one  barium  radical 


IO2 — 

and  will  exactly  unite  with  it  to  form  a  molecule.  The  same 
with  the  nitrate.  Sulphate  has  a  power  of  two  and  is  equiva- 
lent to  the  barium  radical  at  once.  The  phosphate  radical  has 
a  power  of  three.  The  least  common  multiple  of  the  two 
powers  concerned,  two  and  three,  is  six.  It  will  take  three 
bariums  to  equal  six ;  and  two  phosphates. 

A1C13,  Aluminum  chloride. 
A1(NO3)3,  Aluminum  nitrate. 
A12(SO4)3,  Aluminum  sulphate. 
A1PO4,  Aluminum  phosphate. 
Al4(Fe(CN)6)3,  Aluminum  ferrocyanide. 

In  the  last  case,  the  two  combining  powers  concerned  are 
three  and  four;  the  least  common  multiple  of  these  two  is 
twelve ;  three  goes  in  twelve  four  times ;  there  are  therefore 
four  aluminums  to  be  used ;  four  goes  in  twelve  three  times, 
which  calls  for  three  ferrocyanide  radicals. 

55.  An  acid  is  a  substance  that  is  generally  sour  to  the  taste, 
turns  blue  litmus  paper  red  and   neutralizes  a  base  to   form 
water  and  a  salt.     It  always  contains  hydrogen  combined  with 
an  acid  radical. 

56.  A  base  is  a  compound  containing  the  radical  OH,  and 
neutralizes   an   acid   forming   water  and   a   salt.     It   generally 
consists  of  a  basic  radical  joined  to  OH. 

57.  An  alkali  is  a  strong  caustic  base  which  is  soapy  to  the 
taste  and  feeling  and  will  turn  red  litmus  blue.     The  common 
alkalis  are  sodium,  potassium,  and  ammonium  hydroxides. 

58.  A  salt  is  a  compound  of  any  basic  radical  except  H, 
with  any  acid  radical  except  O  or  OH,  and  is  always  formed 
together  with  water  when  an  acid  combines  with  a  base  or  a 
basic  oxide. 

Exercise. 

59.  Give  the  names  of  the  compounds  of  the  basic  radicals 
with  the   acid,   stating   in   each   case   whether   the   compound 
is  an  acid,  a  base,  or  a  salt. 

60.  A  basic  oxide  is  one  that  unites  with  an  acid  to  form 
water  and  a  salt.     Most  of  the  oxides  of  the  metals  are  basic 
oxides. 

MgO;  ZnO;  HgO;  PbO;  CuO. 


—103— 

61.  An  acidic  oxide  is  one  that  unites  with  a  base  to  form 
water  and  a  salt.     Most  of  the  oxides  of  the  non-metals  are 
acidic  oxides. 

CO2;  SiO2;  SO2;  SO3;  P2O5. 

62.  An  alkaline  oxide  is  a  basic  oxide  that  unites  with  water 
to  form  an  alkali. 

Na2O;   K2O;   CaO;   BaO;   MgO. 

63.  An  acid  anhydride  is  an  acidic  oxide  that  unites  with 
water  to  form  an  acid. 

C02;  SO.;  SO3;  P2O3;  P,O5- 


REACTIONS  RESULTING  IN  THE  FORMA- 
TION OF  SALTS. 

643.     THE    ACTION    OF    A    METAL    ON    AN    ACID. 

In  this  reaction  the  metal  takes  the  place  of  an  equivalent 
quantity  of  hydrogen  in  the  acid,  and  the  hydrogen  is  set  free. 
In  case  there  is  an  oxidizing  agent  present,  the  hydrogen  would 
be  oxidized  to  water  and  would  not  appear  in  the  free  con- 
dition. 

2Na  +  2HC1  =  2NaCl  +  H2. 

Zn  +  2HC1  =  ZnCl2  +  H2. 

Zn  +  H2SO4  =  ZnSO4  +  H2. 

2A1  +  3H2S04  =  A12(S04)3  +  3H2. 

Combine  each  of  the  following  basic  radicals  with  each  of 
the  acid  radicals,  and  write  equations  indicating  the  for- 
mation of  the  resulting  salts,  by  the  above  method. 

Basic  radicals.  Acid  radicals. 

Na/  Cl.' 

K/  NO./ 

Ba."  S04." 

Mg."  P04/" 

Al/"  Fe(CN)6."" 

b.     THE  ACTION  OF  A  BASIC  OXIDE  ON  AN  ACID. 

In  this  case  the  oxygen  of  the  oxide  unites  with  the  hydro- 
gen of  the  acid,  forming  water,  thus  leaving  the  basic  radical  of 
the  oxide  free  to  combine  with  the  acid  radical  of  the  acid. 

Na2O  +  2HC1  =  2NaCl  +  H0O. 

ZnO  +  2HC1  =  ZnCl2  +  H26. 

AL03  +  3H2S04  =  A12(S04)3  +  3H20. 

Write  equations  showing  the  formation  of  the  same  twenty- 
five  salts  by  this  method. 


—105— 


c.     THE  ACTION  OF  A  BASE  ON  AN  ACID. 

In  this  case,  the  hydroxide  of  the  base  unites  with  the  hydro- 
gen of  the  acid,  forming  water,  thereby  leaving  the  basic 
radical  of  the  base  free  to  combine  with  the  acid  radical 
of  the  acid. 

NaOH  +  HC1  =  NaCl  +  HOH. 

Zn(OH)2  +  2HC1  =  ZnCl2  +  2HOH. 

2A1(OH)3  +  3H2S04  =  A12(S04)3  +  6HOH. 

Write  equations  showing  the  formation  of  the  same  twenty- 
five  salts  by  this  method. 


d.     THE  ACTION  OF  AN  ACID  ON  THE  SALT  OF 
A  MORE  VOLATILE  ACID. 

In  this  reaction  the  hydrogen  of  the  acid  unites  with  the 
acid  radical  of  the  salt  and  forms  another  acid,  which  is 
either  unstable  and  breaks  up  into  volatile  gaseous  compounds 
at  once,  or  evaporates  as  an  acid  either  with  or  without  the 
application  of  heat,  leaving  the  acid  radical  of  the  original 
acid  combined  with  the  basic  radical  of  the  salt. 

Carbonic  acid,  or  hydrogen  carbonate,  H2CO3,  is  one  of  the 
most  unstable  and  volatile  acids,  therefore  almost  any  acid  will 
decompose  a  carbonate.  Sulphuric  and  phosphoric  acids  are 
very  stable  and  non-volatile.  Therefore,  when  almost  any 
salt  is  treated  with  sulphuric  acid,  another  acid  is  set  free  and 
a  sulphate  is  formed. 

Na2CO3  +  2HC1  =  2NaCl  +  H2O  +  CO2. 


ZnCO3  +  2HC1  =  ZnCl,  +  Ho 
2A1(N03)3  +  3H2S04  =  A12(S04)8  +  6HN03. 

Write  equations  showing  the  formation  of  the  same  twenty- 
five  salts  by  this  method,  making  use  of  the  carbonates  contain- 
ing the  required  basic  radical. 

In  order  that  the  student  may  intelligently  make  use  of  this 
kind  of  reaction,  it  is  important  that  he  should  have  some  idea 
of  the  relative  degrees  of  volatility  or  stability  of  some  of  the 


— 106 — 

more  common  acids.  For  this  purpose  all  acids  might  be 
divided  into  three  classes,  those  that  are  extremely  non-volatile 
and  stable,  those  that  are  easily  volatile  but  of  sufficient  sta- 
bility to  admit  of  being  prepared  in  the  free  condition,  and  those 
that  are  so  unstable  that  they  decompose,  the  moment  they  are 
formed,  into  water  and  oxides  of  non-metals. 

Examples  of  acids  that  are  non-volatile  and  stable. 

H2SO4,  Hydrogen  sulphate  or  sulphuric  acid. 

H3PO4,  Hydrogen  ortho-phosphate  or  ortho-phosphoric  acid. 

Examples  of  acids  that  are  volatile  and  stable. 

HNO3,  Hydrogen  nitrate  or  nitric  acid. 
HC1,  Hydrogen  chloride  or  hydrochloric  acid. 
HC2H3O2,  Hydrogen  acetate  or  acetic  acid. 

Examples  of  acids  that  are  unstable. 

H2CO3,  Hydrogen  carbonate  or  carbonic  acid. 
H2SO3,  Hydrogen  sulphite  or  sulphurous  acid. 
HNO2,  Hydrogen  nitrite  or  nitrous  acid. 


e.  THE  MIXING  OF  TWO  SOLUTIONS,  ONE  OF 
WHICH  CONTAINS  THE  BASIC  RADICAL  AND 
THE  OTHER  THE  ACID  RADICAL  OF  AN 
INSOLUBLE  COMPOUND. 

In  this  case  the  insoluble  compound  separates  as  a  precipitate 
at  once,  and  may  be  filtered  off.  If  the  two  substances  have 
been  mixed  in  chemically  equivalent  proportions,  the  filtrate 
may  be  evaporated  and  the  other  compound  obtained.  The 
insoluble  compound  may  be  an  acid,  a  base,  or  a  salt.  In  mak- 
ing use  of  this  method,  it  is  necessary  to  know  whether  or  not 
a  substance  is  soluble.  For  this  purpose  a  table  of  solubilities 
may  best  be  consulted.  165. 

Silver  chloride,  barium  sulphate,  and  calcium  carbonate  are 
almost  completely  insoluble  in  water ;  some  one  of  them  may  be 
used  in  preparing  any  salt  by  this  method.  It  is  also  well  to 
remember  that 


—ID;— 

ALL  THE  HYDROGEN,  SODIUM,  POTASSIUM 
AND  AMMONIUM  COMPOUNDS  ARE  SOLUBLE, 
ALSO  THE  CHLORATES,  ACETATES  AND  NI- 
TRATES AND  ALL  THE  CHLORIDES  EXCEPT 
THOSE  OF  SILVER,  LEAD,  AND  MERCUROUS 
MERCURY. 

In  writing  equations  showing  the  formation  of  insoluble 
compounds,  the  latter  should  be  indicated  by  a  line  drawn  under- 
neath the  formula:  e.g.  It  is  desired  to  show  the  formation 
of  sodium  chloride  by  this  method. 

Making  use  of  the  insoluble  compound,  barium  sulphate, 
we  see  that  we  must  have  two  soluble  compounds  which  by 
their  combination  will  form  sodium  chloride  and  barium  sul- 
phate ;  these  two  compounds  must  therefore  be  barium  chloride 
and  sodium  sulphate.  We  know  these  are  both  soluble  in  water, 
because,  according  to  the  rule,  one  is  a  chloride  and  the  other 
a  sodium  compound.  We  therefore  have 

Na2S04  +  BaCl2  =  2NaCl  +  BaSO4. 

ZnSO4  +  BaCl2  =  ZnCl2  +  BaSO4. 

2A1C13  +  3Ag2S04  =  A12(S04)3  +  6AgCl. 

Show,  by  equations,  the  formation  of  the  same  twenty-five 
salts,  using  in  each  case  one  of  the  above  three  insoluble 
compounds. 

f.     THE  DIRECT  UNION  OF  A  BASIC  OXIDE  WITH 
AN  ACIDIC  OXIDE. 

This  method  applies  only  to  the  formation  of  oxygen  salts. 
Na2O  +  SO3  =  Na2SO4. 


PbO  +  SiO2  =  PbSiO3. 

Combine  each  of  the  following  acidic  oxides  with  each  of 
the  basic  oxides. 

Acidic  oxides.  Basic  oxides. 

N2O5    forms  nitrates.  Na2O. 

P2O5     forms  phosphates.  K2O. 

As2O3  forms  arsenites.  BaO. 

SO3      forms  sulphates.  A12O3. 


— io8— 

Acidic  oxides.  Basic  oxides. 

CrO3     forms  chromates.      PbO. 
SiO2      forms  silicates.          Fe2O3. 

New  radicals.  AsO3'",  Arsenite.  CrO4",  Chromate. 
SiO4"",  Orthosilicate.  NO2,  Nitrite.  SO3,  Sulphite. 

g.     THE  ACTION  OF  AN  ACID  ANHYDRIDE  ON  A 

BASE. 

In  this  case  water  and  a  salt  are  formed.  The  action  is 
essentially  the  same  as  the  action  of  an  acid  on  a  base,  for  the 
acid  anhydride  acting  on  the  water  at  once  forms  the  corre- 
sponding acid. 

Write  equations  showing  the  combination  of  the  following 
acid  anhydrides  with  the  indicated  bases : 

Acid  anhydrides.  Bases. 

CO2.  NH4OH. 

SiO2.  NaOH. 

N2O3.  KOH. 

502.  Ca(OH)2. 

503.  Ba(OH)2. 
Cr03.  Sr(OH)2. 

In  writing  equations  illustrating  this  method,  it  will  be  best 
for  the  student  at  first  to  write  an  equation  showing  the  forma- 
tion of  the  corresponding  acid  by  the  union  of  the  anhydride 
with  water,  then  write  a  second  equation  showing  the  combina- 
tion of  the  base  with  the  quantity  of  acid  formed.  Adding 
these  two  equations  algebraically  gives  the  required  final 
question. 

Thus, 

N2O3  +  H2O  =  2HNO2. 

2HND2  +  2NaOH  =  2NaNO2  +  2H2O. 

N2O3  +  2NaOH  —  2NaNO2  +  H2O. 
or, 

P205  +  3H20  =  2H3P04. 

2H3PO4  +  3Ca(OH)2  =  Ca3(PO4)2  +  6H2O. 

P205  +  3Ca(OH)2  =  Ca3(P04)2  +  3H2O. 
It  ought  not  to  be  necessary  to  write  the  two  preliminary 
equations, 

SO3  +  Ca(OH)2  =  CaSO4  +  H2O. 


OXIDATION. 

65.  Most  of  the  elements  combine  directly  or  indirectly  with 
oxygen  to  form  oxides,  and  many  of  them  combine  with  differ- 
ent quantities  of  oxygen,   forming  different  oxides.     In  such 
cases  we  recognize  what  may  be  called  the  stage  of  oxidation. 
An  element  may  exist  in   its  highest,  lowest  or-  some  inter- 
mediate stage  of  oxidation. 

Oxidation  is  the  process  of  adding  oxygen  to  or  extracting 
hydrogen  from  a  substance.  It  is  also  the  process  of  chang- 
ing an  element  from  a  lower  to  a  higher  stage  of  oxidation. 
Thus  we  might  consider  chlorine  in  the  compound  HC1  to  be  in 
its  lowest  stage  of  oxidation.  If  this  compound  is  oxidized 
chlorine  is  set  free.  Free  chlorine  would  then  represent  a 
higher  stage  of  oxidation  than  when  combined  with  hydrogen. 

66.  An  oxidizing  agent  is  any  substance  that  directly  or 
indirectly  brings  about  oxidation. 

67.  Reduction    is    the    process    of    extracting    oxygen    or 
adding  hydrogen   to   a   substance.     It  is   also   the  process   of 
changing   an    element    from    a   higher    to    a    lower    stage    of 
oxidation. 

All  oxidizing  processes  may  be  divided  into  two  classes, 
oxidation  in  the  dry  way  and  oxidation  in  the  wet  way. 

Oxidation  in  the  Dry  Way. 

68.  In  cases  of  oxidation  in  the  dry  way  the  oxidizing  agent 
is  commonly  free  gaseous  oxygen  or  some  oxygen  compound 
which  readily  gives  up  some  or  all  of  its  oxygen.     Such  reac- 
tions   take    place    without    the    intervention    of    a    liquid    and 
commonly  therefore  result  in  temperatures  sufficiently  high  to 
produce  light  as  well  as  heat. 

Many  elements  when  heated  in  oxygen  or  in  air  combine 
directly  with  oxygen,  forming  oxides.  If  an  excess  of  oxygen 
is  present  the  highest  stage  of  oxidation  will  ordinarily  be 
attained ;  if  an  excess  of  the  element  is  present  a  lower  form  of 
oxide  may  be  produced.  Some  elements  unite  so  energetically 
with  oxygen  that  the  oxidation  proceeds  by  itself  with  the  pro- 
duction of  light  as  well  as  heat.  Such  substances  are  said  to 
burn.  Magnesium,  sulphur  and  carbon  burn  in  air.  Chemi- 


—no — 

cal  combination  accompanied  by  light  and  heat  is  called  combus- 
tion. Burning  is  simply  a  common  form  of  combustion. 

If  a  substance  consisting  of  several  combustible  elements  is 
ignited  in  air  it  generally  burns,  and  the  products  of  the  burning 
are  the  oxides  of  the  elements  it  contained.  Wood,  illumi- 
nating gas,  oil  and  alcohol  are  examples  of  such  substances. 

Metallic  oxides  themselves  are  oxidizing  agents  if  heated  with 
some  material  that  combines  very  readily  with  oxygen. 
Examples  of  materials  that  have  great  power  of  uniting  with 
oxygen  are  hydrogen,  carbon,  sodium,  potassium,  aluminum 
and  magnesium. 

C  =  2Pb  +  CO2. 
+  3C  =  4Fe  +  3CO2. 
CuO  +  H2  =  Cu  +  H20. 

It  should  be  noted  that  oxidation  is  always  accompanied  by 
reduction.  The  oxidizing  agent  oxidizes  the  reducing  agent 
and  the  reducing  agent  reduces  the  oxidizing  agent. 

Some  Oxidizing  Agents  for  the  Dry  Way: 

Oxygen. 

Air. 

KC1O3,  Potassium  chlorate. 

KNO3,  Potassium  nitrate. 

Pb3O4,  Red  lead. 

MnO2,  Manganese  dioxide. 

Equations  illustrating  oxidation  in  the  dry  way,  where 
oxygen  is  the  oxidizing  agent. 

C  +  02  =  C02. 

2Cu  +  O2  =  2CuO. 

4Cu  +  O2  =  2Cu2O. 

2Mg  +  O2  =  2MgO. 

2C2H6  +  7<32  =  4CO2  +  6H0O. 

C2H5OH  +  302  =  2C02  +  3H20. 

Exercise. 

Write  equations  illustrating  the  oxidation  or  burning  of 
the  following  substances : 

Sodium,  to  form  sodium  oxide. 
Magnesium,  to  form  magnesium  oxide. 


— Ill — 

Zinc,  to  form  zinc  oxide. 
Aluminum,  to  form  aluminum  oxide. 
Carbon,  to  form  CO2. 
Nitrogen,  to  form  N2O3. 
Nitrogen,  to  form  N2O5. 
Phosphorus,  to  form  P2O3. 
Phosphorus,  to  form  P2O5. 
Sulphur,  to  form  SO2. 
Sulphur,  to  form  SO3. 

69.  The  Oxides  of  the  Elements  in  Groups  in  Relation 
to  Acids  and  Bases. 

All  the  elements  may  be  divided  into  two  classes,  metals  and 
non-metals.  In  general,  the  oxides  of  the  metals  unite  directly 
or  indirectly  with  water  to  form  bases.  The  oxides  of  the  non- 
metals  unite  with  water  either  directly  or  indirectly  to  form 
acids. 

The  First  Group. 

All  the  elements  of  this  group  are  metals  and  all  form  oxides 
of  the  general  formula  R2O.  Copper,  in  addition  to  the  oxide 
Cu2O,  forms  the  oxide  CuO.  These  oxides  are  all  base  pro- 
ducing. 

Na2O  +  H2O  =  2NaOH. 

The  Second  Group. 

All  the  elements  of  this  group  are  metals  and  form  oxides 
of  the  general  formula  RO.  Mercury,  in  addition  to  the  oxide 
HgO,  forms  the  oxide  Hg2O.  These  oxides  are  also  all  base 
producing. 

CaO  +  H2O  =  Ca(OH)2. 

The  Third  Group. 

Two  elements  only  are  considered  in  this  group,  boron  and 
aluminum.  Aluminum  forms  the  oxide  A12O3,  which  reacts 
with  water  indirectly  to  form  the  compound  A1(OH)3,  which 
acts  as  a  base  toward  strong  acids  and  as  an  acid  toward  strong 
bases. 

Boron  is  a  non-metal  and  forms  the  oxide  B2O3,  which  com- 
bined with  water  forms  the  compound  B(OH)3.  This  com- 


— 112 — 

pound  has  only  acid  properties  and  therefore  the  formula  is 
usually  written  H3BO3  and  is  known  as  boric  acid. 

The  Fourth  Group. 

Carbon  forms  two  oxides.  The  first,  CO,  forms  neither  acids 
nor  bases.  The  second,  CO2,  unites  with  water  to  form  an 
unstable  acid,  CO(OH)2,  called  carbonic  acid  and  commonly 
written  H2CO3. 

Silicon   forms  one  oxide,   SiO2,  which   forms  a  number  of 
acids  when  combined  with  varying  amounts  of  water. 
SiO2  +  H2O  =  SiO(OH)2  or  H2SiO3,  called  meta-silicic  acid. 
SiO2  +  2H2O  =  Si(OH)4  or  HJSiO4,  called  ortho-silicic  acid. 

Tin  forms  two  oxides,  SnO,  basic,  and  SnO2,  both  basic  and 
acidic. 

SnO2  +  H2O  =  SnO(OH)2  or  H2SnO3,  stannic  acid. 

Lead  forms  several  oxides,  only  one  of  which,  PbO,  need  be 
considered  here.  This  oxide  unites  with  water  to  form 
Pb(OH)2,  which  is  basic  toward  acids  and  weakly  acidic  toward 
strong  bases. 

The  Fifth  Group. 

Nitrogen  forms  several  oxides,  three  of  which  are  acid 
producers. 

N2O   +  H2O  =  2HNO,    Hyponitrous  acid. 
N2O8  +  H2O  =  2HNO2,  Nitrous  acid. 
N2OB  +  H2O  =  2HNO~,  Nitric  acid. 

Phosphorus  forms  two  oxides  each  of  which  forms  acids 
with  water. 

P2O8  +  3H2O  =  2P(OH)3  or  2H3PO8,  phosphorus  acid. 

P2O5  +  H2O  =  2liPO3,  meta-phosphoric  acid. 

P2O5  +  2H2O  =  H4P2O7,  pyro-phosphoric  acid. 

P2O5  +  3H2O  =  2H3PO4,  ortho-phosphoric  acid. 

Arsenic  and  antimony  form  oxides  and  acids  similar  to  those 
of  phosphorus. 

Bismuth  forms  several  oxides,  only  one  of  which  need  be 
considered  here. 

Bi2O3  is  a  basic  oxide  forming  bismuth  hydroxide  with  water. 


The  Sixth  Group. 

Sulphur  forms  two  oxides,  SO2  and  SO3,  both  acidic. 

502  +  H2O  =  H2SO3,  sulphurous  acid. 

503  +  H2O  =  H2SO4,  sulphuric  acid. 

Chromium  does  not  form  the  oxide  CrO  but  the  correspond- 
ing hydroxide,  Cr(OH)2,  is  well  known.  Cr2O3,  chromic  oxide, 
is  weakly  basic  and  forms  chromic  hydroxide,  Cr(OH)3. 

CrO3,  chromium  trioxide  or  chromic  anhydride,  is  acidic  and 
forms  chromic  acid,  CrO(OH)2  or  H2CrO4. 

The  Seventh  Group. 

This  group  with  the  exception  of  manganese  may  best  be 
considered  as  forming  a  regular  series  of  acids  corresponding 
to  the  four  regular  valencies.  The  exceptions  will  be  learned 
in  the  study  of  the  separate  elements. 

C12O   -f  H2O  =  2HC1O,    hypochlorous  acid. 
C12O3  +  H2O  =  2HC1O2,  chlorous  acid. 
C12O5  +  H2O  =  2HC1O3,  chloric  acid. 
C12O7  +  H2O  =  2HC1O4,  per-chloric  acid. 

Bromine  does  not  form  any  oxides  but  the  acids  HBrO, 
hypobromous,  and  HBrO3,  bromic,  are  known.  Iodine  may  be 
considered  as  forming  acids  similar  to  the  chlorine  acids.  Only 
one  oxide  has  been  prepared  in  the  free  condition,  I2O5. 

Manganese  forms  six  oxides,  some  of  which  are  basic  and 
some  acidic. 

MnO,  manganous  oxide,  basic,  forming  Mn(OH)2. 
Mn2O3,  manganic  oxide,  forming  Mn(OH)3,  basic. 
MnO3,  manganese  trioxide  or  manganic  anhydride. 
MnO3  -f-  H2O  =  H2MnO4,  manganic  acid. 
Mn2O7,  manganese  hept-oxide  or  per-manganic  anhydride. 
Mn2O7  +  H2O  =  2HMnO4,  per-manganic  acid. 

The  Eighth  Group. 

The  important  oxides  of  this  group  correspond  to  the 
valencies  of  two  and  three.  They  are  all  basic  and  form  basic 
hydroxides. 


—114— 

70.  The  System  of  naming  Acids  and  Salts. 

The  names  of  salts  formed  from  "ous"  acids  end  in  "ite," 
and  the  names  of  salts  formed  from  "ic"  acids  end  in  "ate." 

If  an  acid  has  less  oxygen  than  the  usual  "ous"  acid,  that  is 
if  it  represents  a  lower  stage  of  oxidation,  the  prefix  "hypo" 
is  placed  before  the  name  of  the  "ous"  acid.  Thus  HNO2  is 
nitrous  acid,  HNO  is  hyponitrous  acid. 

If  an  acid  represents  a  higher  stage  of  oxidation  than  the 
usual  "ic"  acid,  the  prefix  "per"  is  placed  before  the  name. 
Thus,  HC1O3  is  chloric  acid  and  HC1O4  is  per-chloric  acid. 

The  Names  and  Valencies  of  Radicals. 

The  names  of  radicals  are  taken  as  the  first  or  last  part  of  the 
name  of  a  salt.  Thus, 

NaNO  is  sodium  hypo-nitrite,  NO  is  the  hypo-nitrite  radical 
and  has  a  valence  of  one  because  it  can  form  a  compound  with 
one  hydrogen. 

Mg2P2O7  is  magnesium  pyro-phosphate,  P2O7  is  the  pyro- 
phosphate  radical  and  has  a  valence  of  four  because  it  can  exist 
in  combination  with  four  hydrogen  atoms. 

OXIDATION  IN  THE  WET  WAY. 

71.  Many  oxidations  can  best  be  carried  out  in  a  solution 
containing  the  required  oxidizing  agent  and  the  substance  to  be 
oxidized. 

In  such  reactions  no  oxygen  is  actually  set  free  in  the  gaseous 
form  but  enters  at  once  into  combination  with  the  reducing 
agent.  If  there  is  an  excess  of  the  reducing  agent  all  the 
oxidizing  agent  will  be  used  up.  If  there  is  an  excess  of  the 
oxidizer,  all  the  reducing  agent  will  be  transformed.  If  any- 
basic  oxides  are  formed  in  the  reaction  an  acid  must  be  present 
to  change  them  to  the  corresponding  salts. 

Chlorine,  bromine  and  iodine  oxidize  in  the  presence  of  water 
by  virtue  of  extracting  the  hydrogen  of  the  water,  forming  the 
corresponding  acid,  and  liberating  oxygen. 

2C12  +  2H2O  =  4HC1  +  O2. 
2Br2  +  2H2O  =  4HBr  +  O2. 
2l2  +  2H2O  =  4HI  +  Oo, 


—US- 
Compounds  containing  oxygen  and  two  other  elements  may 
be  considered  as  made  up  of  two  oxides.  If  the  compound  is 
an  acid,  one  of  the  oxides  will  be  water  and  the  other  an  acid 
anhydride.  If  it  is  a  salt,  one  of  them  will  be  a  basic  oxide 
and  the  other  an  acid  anhydride. 

In  either  case,  if  the  acid  anhydride  is  an  oxide  of  an  element 
corresponding  to  a  high  combining  power,  and  if  an  oxide  corre- 
sponding to  a  lower  combining  power  exists,  the  substance  is  an 
oxidizing  agent. 

Consider  sulphuric  acid,  H2SO4. 

H2SO4  —  H2O  +  SO3,  but  SO3  can  give  up  oxygen 
and  drop  to  SO2.     Therefore  we  have 

H2S04  =  H20  +  S02  +  O. 
2K2CrO4  =  2K,O  +  2CrO3. 

CrO3  can  give  up  oxygen  and  become  Cr2O3 ;   so  we  have 
2K2Cr04  =  2K20  +  Cr203  +  3O. 

2KMnO4  =  K2O  +Mn2O7. 
Mn2O7  can  drop  to  2MnO ;  therefore  we  have, 
2KMnO4  =  K2O  +  2MnO  +  50. 

Exercise. 

72.  State  the  formulas  of  ten  different  oxidizing  agents  and 
indicate,  by  equations,  how  they  oxidize. 

Formation   of   Equations   for   Oxidation   Reactions   in   the 

Wet  Way. 

73.  It  is  required  to  oxidize  phosphorus  with  nitric  acid. 
First,  how  does  nitric  acid  oxidize? 

2HNO3  =  H2O +2NO +3O,  that  is,  when  nitric  acid  is 
used  as  an  oxidizing  agent,  two  molecules  give  three  atoms  of 
oxygen,  and  nitric  oxide  is  given  off. 

When  phosphorus  is  oxidized,  two  atoms  require  five  atoms 
of  oxygen  to  form  the  pentoxide.  Oxygen  is  furnished  by  the 
nitric  acid  in  multiples  of  three  and  needed  by  the  phosphorus 
in  multiples  of  five.  Therefore  we  shall  have  to  use  five  times 


—  no— 

two  molecules  of  nitric  acid  and  three  times  two  atoms  of 
phosphorus. 

ioHNO3  =  sH2O  +  loNO  +  150. 


Phosphorus  pentoxide  is  an  acid  anhydride  and  will  combine 
with  the  proper  quantity  of  water  to  form  phosphoric  acid. 

3P205  +  9H20  =  6H.PO.. 

Adding  these  equations  together,  we  have  the  final  equation 
for  this  reaction.  • 

4H2O  +  6P  +  ioHNO8  =  6H3PO4  +  icNO. 

Exercise. 

74.  Indicate  the  oxidation  of  the  other  elements  of  the  fifth 
group,  in  the  same  way. 

75.  If  the  oxidizing  agent  is  a  substance  that  gives  rise  to  a 
basic  oxide  in  decomposing,  there  must  be  an  acid  present  to 
combine  with  this  oxide  and  form  the  corresponding  salt. 

Consider  manganese  dioxide  with  hydrochloric  acid. 

MnO2  =  MnO  +  O.     MnO  is  a  basic  oxide. 
One  atom  of  O  will  combine  with  two  atoms  of  H. 

O  +  2HC1  =  H20  +  C12. 
The  MnO  requires  two  molecules  of  hydrochloric  acid. 

MnO  +  2HC1  =  MnCl2  +  H2O. 
Adding  these  equations  together,  we  have 

MnO2  +  4HC1  =  MnCl2  +  2H2O  +  C12. 

Exercise. 

76.  Indicate  the  oxidation  of  the  other  halogen  acids  in  the 
same  way.     Also  the  oxidation  of  H0SO3,  sulphurous  acid,  to 
H2S04. 

Suppose  we  had  to  oxidize  hydrochloric  acid  with  potassium 
permanganate,  KMnO4. 

2KMnO4  =  K2O  +  2MnO  +  50. 
5O  take  10  H 

50  +  loHCl  =  5H20  +  5C12. 


K2O  takes  2  HC1 

K20  +  2HC1  =  2KC1  +H2O. 
2MnO  take3  4  HC1 

2MnO  +  4HC1  =  2MnCl2  +  2H2O. 

Adding  these  equations  together,  we  have  the  final  equation: 
2KMnO4  +  i6HCl  =  2MnCl2  +  2KC1  +  8H2O  +  sC!2. 


EXAMPLES  OF  OXIDATION  INVOLVING  A 
CHANGE  FROM  A  LOWER  TO  A  HIGHER 
STAGE  OF  OXIDATION. 

77.  Change  cuprous  chloride,  CuCl,  to  cupric  chloride, 
CuCl2.  Add  chlorine  directly, 

CuCl  +  Cl  =  CuCl2. 

Or  add  hydrochloric  acid  and  an  oxidizing  agent.  Use  nitric 
acid.  Two  molecules  of  this  acid  give  three  atoms  of  oxygen 
for  oxidizing  purposes;  three  atoms  of  oxygen  are  equivalent 
to  six  atoms  of  hydrogen  or  to  six  molecules  of  hydrochloric 
acid.  So  the  least  quantity  of  chlorine  that  we  can  use  will  be 
six  atoms.  We  must  therefore  oxidize  six  molecules  of  cuprous 
chloride. 

Writing  all  the  equations  and  adding  them  together,  we  have 

2HNO3  =  H2O  +  2NO  +  30. 
30  +  6HC1  =  3H20  +  3C12. 
3C12  +  6CuCl  =  6CuCl2. 

6CuCl  +  2HN03  +  6HC1  =  6CuCl2  +  2NO  +  4H2O. 

Indicate  this  same  oxidation,  using  different  oxidizing  agents. 

Suppose  it  is  required  to  oxidize  ferrous  sulphate  to  ferric 
sulphate  by  means  of  potassium  permanganate. 

The  formula  of  ferrous  sulphate  is  FeSO4,  and  for  ferric 
sulphate  is  Fe2(SO4)s.  It  is  evident  that  two  molecules  of 
the  ferrous  salt  will  be  required  to  make  one  of  the  ferric. 

The  basic  oxide  contained  in  the  ferrous  sulphate  is  FeO, 
and  in  the  ferric  sulphate  is  Fe2O3.  For  each  two  molecules 
of  the  first  compound  therefore  there  must  be  one  atom  of 
oxygen  added,  as  well  as  some  more  SO4. 


— n8— 

Potassium  permanganate  oxidizes  in  acid  solution  according 
to  the  equation, 

2KMnO4  =  K2O  +  2MnO  +  50. 

Since  five  oxygen  atoms  are  available,  at  least  ten  molecules 
of  ferrous  sulphate  must  be  used,  in  which  case  five  molecules 
of  sulphuric  acid  will  be  necessary  to  furnish  the  five  additional 
SO4  radicals. 
ioFeSO4  +  2KMnO4  +  sH2SO4  = 

5Fe2(S04)3  +  K20  +  2MnO  +  5H2O. 
2MnO  +  2H2SO4  ==  2MnSO4  +  2H2O. 

K20+    H2S04  =     K2S04+    H20. 

ioFeSO4  +  2KMnO4  +8H2SO4  = 

5Fe2(S04)3  +  2MnS04  +  K2SO4  +  8H2O. 

SOME  OXIDIZING  AGENTS  WITH  EQUATIONS 
SHOWING  HOW  THEY  OXIDIZE  IN  THE  WET 
WAY. 

78.     i.  PbO2  =  PbO  +  O. 

2.  MnO2  =  MnO  +  O,  in  the  presence  of  a  reducing  agent 
and  an  acid. 

3.  C12  +  H2O  •=  2HC1  +  O,  in  the  presence  of  a  reducing 
agent. 

4.  Br2  +  H2O  =  2HBr  +  O,  in  the  presence  of  a  reducing 
agent. 

5.  I2  +  H2O  =  2HI  -f-  O,  in  the  presence  of  a  reducing  agent. 

6.  KC1O3  =  KC1  -|-  3O,  when  heated  to  a  high  temperature. 

7.  Pb3O4  =  3?bO  +  O,  when  heated  or  in  the  presence  of 
an  acid  and  a  reducing  agent. 

8.  2HNO3  =  H2O  +  2NO  +  30,  in  the  presence  of  a  reduc- 
ing agent. 

9.  K2CrO4,  potassium  chromate,  2K2CrO4  =  2K2O  +  Cr2O3 
+  3O,  in  the  presence  of  an  acid  and  a  reducing  agent. 

10.  K2Cr2O7,    potassium    bichromate,    dichromate    or    acid 
chromate.     K2Cr2O7  =  K2O  +  Cr2O3  +  3O,  in  the  presence  of 
an  acid  and  a  reducing  agent. 

11.  KMnO4,   potassium   permanganate,    2KMnO4  =  K2O  + 
2MnO  -f~  5O,  in  the  presence  of  a  reducing  agent  and  an  acid. 

12.  K2MnO4,    potassium    manganate.       K2MnO4  =  K2O  -\- 
MnO  +  2O,  in  the  presence  of  an  acid  and  a  reducing  agent. 


Exercise. 

79.     Indicate,  by  equations,  the  following  oxidations,  making 
use  in  each  case  of  the  above  oxidizing  agents. 
C,  carbon,  to  CO2,  carbon  dioxide.     68. 
CO,  carbon  monoxide,  to  CO2,  carbon  dioxide. 
H,  hydrogen,  to  H2O,  water  or  hydrogen  oxide. 
CH4,  methane,  to  CO2  +  H2O. 
C2H5OH,  alcohol,  to  CO2  +  H2O. 
CH5OH  to  CH3CHO,  aldehyde. 
C2H5OH  to  CH3COOH,  acetic  acid. 
S,  sulphur,  to  SO2,  sulphur  dioxide. 
S,  sulphur,  to  SO3,  sulphur  trioxide. 
H2S,  hydrogen  sulphide,  to  S  -f-  H2O. 
H2S,  hydrogen  sulphide,  to  H2SO4. 
HC1,  hydrogen  chloride,  to  H2O  +  Cl.     75. 
HBr,  hydrogen  bromide,  to  H2O  -f  Br.     75. 
HI,  hydrogen  iodide,  to  H2O  +  I.     75. 
HI,  hydrogen  iodide,  to  HIO3,  iodic  acid. 
H2C2O4,  oxalic  acid,  to  CO2  +  H2O. 
C17H35CO2H,  stearic  acid,  to  CO2  +  H2O. 
FeCl2,  ferrous  chloride,  to  FeCl3,  ferric  chloride.     77. 
SnCl2,  stannous  chloride,  to  SnCl4,  stannic  chloride.     77. 
HgCl,  mercurous  chloride,  to  HgCL,  mercuric  chloride.     77. 
FeSO4,  ferrous  sulphate,  to  Fe2(SO4)3,  ferric  sulphate.     77. 
P,  phosphorus,  to  H3PO4,  phosphoric  acid.     73. 
P2O3,  phosphorus  trioxide,  to  H3PO4,  phosphoric  acid. 
As,  arsenic,  to  H3AsO4,  arsenic  acid.     73. 
As2O3,  arsenic  trioxide,  to  H3AsO4,  arsenic  acid. 


REDUCTION  REACTIONS. 

80.  Reduction  is  exactly  the  opposite  of  oxidation.  The 
oxidizing  agent  oxidizes  the  reducing  agent,  and  the  reducing 
agent  reduces  the  oxidizing  agent. 

Any  substance  that  will  combine  directly  with  oxygen  is 
necessarily  a  Reducing  agent;  or  any  substance  that  will  fur- 
nish or  give  up  anything  that  will  combine  with  oxygen. 

Some  common  reducing  agents: 


I2O 

H,  C,  CO,  Na,  Mg,  Al,  Fe,  H2S,  HI,  SO2,  H2SO3,  sulphurous 
acid,  SnQ2,  stannous  chloride,  CH4,  methane,  H2C2O4,  oxalic 
acid,  FeQ2,  ferrous  chloride,  Zn,  FeSO4,  ferrous  sulphate. 

Si.     Examples  of  Reduction  Reactions. 

Fe203  +  3H2  =  2Fe  +  3H2O. 
2FeCl3  +  Fe  =  3FeCl2. 
2FeCl3  +  H2S  =  2FeCl2  +  2HC1  +  S. 
2H2S04  +  C  =  2H20  +  2S02  +  C02. 
4Cu2O  +  CH4  =  8Cu  +  CO2  +  2H2O. 

Exercise. 

82.  Show  the  reduction  of  the  various  metallic  oxides,  each 
with  H,  C  and  CO. 

Reduce  CuCl2  to  CuCl  with  copper. 

Reduce  HgCl2  to  HgCl  with  mercury. 

Reduce  HgCl2  to  HgCl  with  stannous  chloride. 

Reduce  FeCl3  to  FeCl2  with  H2S,  SnCl2  and  H2SO3. 

Reduce  KMnO4  to  K2SO4  and  MnSO4  with  oxalic  acid,  in 
the  presence  of  sulphuric  acid. 

Reduce  K2Cr2O7  to  K2SO4  and  Cr2(SO4)3,  in  the  presence 
of  sulphuric  acid,  with  stearic  acid,  C17H35COOH. 


CHEMICAL   EQUIVALENCE    BETWEEN 
COMPOUNDS. 

83.  One  molecule  of  hydrogen  phosphate,  H3PO4,  is 
equivalent  to  three  molecules  of  hydrochloric  acid,  nitric  acid, 
sodium  hydroxide  or  sodium  chloride. 

Two  molecules  of  hydrogen  phosphate  are  equivalent  to  three 
molecules  of  sulphuric  acid,  calcium  chloride  or  barium 
hydroxide. 

One  molecule  of  sulphuric  acid  is  equivalent  to  two  molecules 
of  hydrochloric  acid,  nitric  acid,  sodium  chloride  or  sodium 
hydroxide. 

One  molecule  of  sulphuric  acid  is  equivalent  to  one  molecule 
of  calcium  nitrate,  zinc  sulphate  or  sodium  carbonate. 

Three  molecules  of  nitric  acid  are  equivalent  to  one  molecule 
of  hydrogen  phosphate;  two  molecules  to  one  molecule  of  sul- 


121 — 

phuric   acid,   and   one   molecule   to   one   molecule   of   sodium 
hydroxide. 

A  certain  quantity  of  one  chemical  compound  is  said  to 
be  equivalent  to  a  definite  quantity  of  another  compound, 
when  they  may  both  be  completely  involved  in  the  same 
reaction  or  can  each  react  with  a  fixed  quantity  of  another 
substance. 

In  general,  the  quantities  represented  in  any  complete  equa- 
tion are  chemically  equivalent  for  that  particular  reaction. 


PART  III. 
MORE  ADVANCED  THEORY. 


84.  Avogadro's  Hypothesis.    Equal  volumes  of  all  gases 
at  the  same  temperature  and  pressure  contain  the  same  number 
of  molecules. 

Or  in  other  words,  all  gaseous  molecules  under  the  same  con- 
ditions occupy  equal  space. 

If  an  equation  has  been  properly  written  so  that  each  symbol 
represents  a  molecule,  then  from  the  above  principle,  the  rela- 
tive volumes  of  the  various  substances  entering  that  reaction, 
when  considered  to  be  in  the  gaseous  form,  are  proportional  to 
the  number  of  molecules  as  indicated  by  the  coefficients  in  the 
equation. 

Example.  How  many  liters  of  oxygen  would  be  required  to 
burn  ten  liters  of  alcohol  vapor? 

C2H5OH  +  302  =  2C02  +  3H20. 

This  equation  means  that  one  molecule  of  alcohol  requires 
three  molecules  of  oxygen  to  burn  it  and  results  in  the  forma- 
tion of  two  molecules  of  carbon  dioxide  and  three  molecules  of 
water. 

In  other  words,  ten  liters  of  alcohol  vapor  require  thirty  liters 
of  oxygen  to  burn  it  and  form  twenty  liters  of  carbon  dioxide 
and  thirty  liters  of  water  vapor. 

Exercise. 

85.  Write  equations  showing  the  oxidation  or  burning  of 
gases  and  show  how  many  liters  of  the  various  products  would 
be  formed  if  ten  liters  of  the  gas  were  used.     68. 


—123— 


MOLECULAR  AND  ATOMIC  WEIGHTS. 

86.  The  hydrogen  molecule  is  divisible.     If  one  volume  of 
hydrogen  be  combined  with  one  volume  of  chlorine,  two  vol- 
umes of  hydrochloric  acid  gas   are   formed.     By  Avogadro's 
Hypothesis  there  must  be  twice  as  many  molecules  of  the  com- 
pound as  there  are  of  either  of  the  elementary  gases.     There- 
fore the  hydrogen  molecule  must  split  up  into  at  least  two 
particles.     The  same  is  true  of  the  chlorine  molecule. 

87.  The  molecule  of  hydrochloric  acid  gas  must  contain 
equal  numbers  of  hydrogen  and  chlorine  atoms.     When  the 
compound  is  analyzed  it  is  found  to  contain  one  part  of  hydro- 
gen to  thirty-five  and  one-half  parts   of  chlorine  by   weight. 
Since  the  atomic  weight  of  hydrogen  is  one  and  that  of  chlorine 
thirty-five  and  one-half,  the  formula  must  be  HnCln. 

88.  The  molecule  of  hydrogen  chloride  contains  but  one 
hydrogen  atom.     If  it  contained  two  or  more  hydrogen  atoms 
a  part  of  this  hydrogen  could  be  replaced  by  a  metal,  forming 
an  acid  salt  having  the  general  formula  NaH^Cln.      Lab.  Ex. 
No.  25.     Or  if  it  contained  two  hydrogen  atoms  a  compound 
having  the  formula  HNaCL  would  have  been  found.     No  such 
compound  has  ever  been  made. 

89.  The  molecule  of  hydrogen  contains  two  atoms.    Since 
it   has   been   demonstrated   that   the    formula   of   hydrochloric 
acid  gas  is  HC1,  and  since  one  volume  or  one  molecule  of  hydro- 
gen gives  two  volumes  or  two  molecules  of  hydrochloric  acid 
gas,  each  molecule  of  hydrogen  must  contain  only  two  atoms. 

90.  Gas  Density  is  the  ratio  of  the  weight  of  a  certain 
volume  of  a  gas  to  the  weight  of  an  equal  volume  of  hydrogen 
at  the  same  temperature  and  pressure.     Sometimes  the  ratio 
of  the  weight  of  a  certain  volume  of  the  gas  to  the  weight  of  an 
equal  volume  of  oxygen  is  taken  and  the  result  multiplied  by 
sixteen. 

91.  Vapor  density  is  an  expression  that  conveys  the  same 
idea  as  gas  density  but  is  more  properly  used  in  reference  to 
substances  that  do  not  ordinarily  exist  in  the  gaseous  form. 

92.  The  molecular  weight  of  a  substance  is  equal  to  twice 
its  gas  density.     The  ratio  of  the  weight  of  a  certain  volume  of 
a  gas  to  the  weight  of  an  equal  volume  of  hydrogen  is  the  same 


—  124— 

as  the  ratio  of  the  weight  of  a  molecule  of  the  gas  to  the  weight 
of  a  hydrogen  molecule,  since  equal  volumes  of  all  gases  under 
the  same  conditions  contain  the  same  number  of  molecules.  In 
other  words,  gas  density  is  a  number  that  tells  how  many  times 
heavier  a  molecule  is  than  one  molecule  of  hydrogen.  Since 
the  molecule  of  hydrogen  contains  two  atoms  and  therefore  has 
a  molecular  weight  of  two,  twice  the  gas  density  must  be  a 
number  that  tells  how  many  times  heavier  a  molecule  of  a  gas 
is  than  one  atom  of  hydrogen.  But  this  is  the  definition  of 
molecular  weight. 

93.     The  determination  of  gas  density.     A  glass  balloon 
similar  to  that  indicated  in  fig.  17  and  having  a  capacity  vary- 


Fig.  17. 

ing  from  100  c.c.  to  1000  c.c.  is  filled  with  a  gas  and  weighed. 
It  is  then  filled  with  hydrogen  and  weighed  again.  These 
weights  are  evidently  not  the  true  weights  of  the  gases,  but  the 
weight  of  the  gas  plus  the  weight  of  the  bulb  minus  the  buoy- 
ant effect  of  the  air  in  which  the  weighing  was  made.  By 
means  of  the  mercury  pump  all  the  gas  is  now  removed  from 
the  bulb,  and  it  is  weighed  again.  This  weight  is  the  weight  of 
the  bulb  less  the  buoyant  effect  of  the  air.  If  this  weight  be 
subtracted  from  the  first  two  weights,  the  real  weights  of  the 
two  gases  will  be  obtained.  By  dividing  the  weight  of  the  gas 
by  the  weight  of  the  hydrogen  the  gas  density  will  be  found 
in  terms  of  hydrogen.  Since  the  ratio  of  the  atomic  wreight 
of  oxygen  to  hydrogen  is  not  exactly  16  to  I,  and  since  tables 
based  on  oxygen  as  16  are  in  common  use,  it  is  better  to  com- 
pare the  weight  of  the  unknown  gas  to  the  weight  of  an  equal 
volume  of  oxygen  and  multiply  the  result  by  16.  This  gives  the 
gas  density  in  terms  of  one-sixteenth  of  the  oxygen  atom. 

94.  Determination  of  vapor  density.  Meyer's  method. 
In  this  method  a  small  quantity  of  the  substance  is  vaporized, 
and  the  volume  of  the  air  displaced  by  the  vapor  is  measured. 


—  125— 


This  volume  is  reduced  to  the  corresponding  volume  at  o°  C. 
temperature  and  760  mm.  pressure  and  multiplied  by  the 
weight  of  i  c.c.  of  oxygen  under  the  same  conditions.  105. 
The  ratio  of  the  weight  of  the  substance  taken,  to  this  weight, 
shows  how  many  times  heavier  the  molecule  is  than  one  mole- 
cule of  oxygen.  This  result  multiplied  by  sixteen  gives  the 
vapor  density  of  the  material.  The  apparatus  used  is  indicated 
in  fig.  1 8.  It  consists  of  two  glass  tubes.  The  outer  one  has 
a  bulb  at  the  bottom  in  which  is  placed  some  liquid  that  has 


-B 


Fig.   18. 

a  higher  boiling  point  than  that  of  the  substance  to  be  vaporized. 
The  vapor  density  tube  is  suspended  in  this  and  the  liquid  in 
the  outer  tube  heated  to  boiling.  After  the  air  in  the  inside 
tube  has  ceased  to  expand,  the  end  of  the  delivery  tube  at  E 
is  placed  under  an  inverted  measuring  tube,  F,  full  of  water. 
About  a  tenth  of  a  gram  of  the  substance  is  placed  in  a  small 
glass  stoppered  vial  and  carefully  weighed.  The  cork  at  D  is 
removed,  the  vial  dropped  in  and  the  cork  quickly  replaced. 
The  vial  passes  through  the  long  inner  tube  to  the  enlarged 


-  126  - 

portion  at  C.  The  heat  causes  the  material  to  vaporize,  the 
stopper  is  blown  out,  and  a  volume  of  air  escapes  into  the 
graduated  tube,  F,  just  equal  to  the  volume  of  the  vapor. 

95.  Determination  of  the  formula  from  the  analysis  and 
vapor  density.  Suppose  a  new  substance  is  found  and  it  is 
desired  to  determine  its  formula.  It  is  first  analyzed  and  its 
chemical  composition  ascertained  in  per  cents.  Then  its  vapor 
density  is  found  by  the  above  methods  or  by  other  methods  to 
be  described  later.  110,  111  and  112. 

If  each  percentage  be  divided  by  the  corresponding  atomic 
weight  a  series  of  numbers  will  be  obtained  which  will  be  pro- 
portional to  the  numbers  of  the  different  atoms  in  the  molecule. 

If  this  series  be  divided  by  the  smallest  number,  the  results 
expressed  as  fractions  and  then  reduced  to  a  common  denomi- 
nator, the  numerators  will  be  the  numbers  of  atoms  of  each 
element  corresponding  to  the  simplest  formula. 

Example.  Suppose  a  substance  contained  carbon,  33.70%  ; 
hydrogen,  3.38%  ;  oxygen,  62.92%  ;  what  would  be  the  for- 
mula corresponding? 

Dividing  these  per  cents,  by  the  corresponding  atomic  weights 
we  have 


.  _ 

This  means  that  the  formula  must  be 

Cs>.81     H3.38     Oa.93. 

Dividing  each  of  these  sub-numbers  by  the  smallest  and 
expressing  the  results  as  fractions,  we  have 

Ci  Hlt  Olf. 

This  means  that  there  are  one  and  one-fifth  as  many  hydrogen 
atoms,  and  one  and  two-fifths  as  many  oxygen  atoms  as  there 
were  carbon  atoms.  This  could  only  be  the  case  if  the  formula 
were 

C5H807. 

If  the  molecular  weight  corresponding  to  this  formula  agrees 
with  that  found  from  the  vapor  density,  then  this  formula  is 
the  correct  one.  Otherwise  it  is  the  simplest  possible  formula, 
and  some  multiple  of  it  will  correspond  to  the  molecular  weight 
called  for  by  the  vapor  density. 


-127- 

If  the  vapor  density  determination  is  exact,  the  molecular 
weight  determined  from  it  may  be  multiplied  at  once  by  the 
various  percentages,  and  the  results  divided  by  the  atomic 
weights  will  give  the  correct  formula. 

Exercise. 

96.  Determine  the  simplest  formulas  corresponding  to  the 
following  analyses : 

1.  Lead,  86.60%,  Sulphur,  13.40%. 

2.  Calcium,  51.10%,  Fluorine,  48.90%. 

3.  Mercury,  84.90%,  Chlorine,  15.10%. 

4.  Arsenic,  75.80%,  Oxygen,  24.20%. 

5.  Hydrogen,  1.59%,  Nitrogen,  22.22%,  Oxygen,  76.19%. 

6.  Iron,  70.00%,  Oxygen,  30.00%. 

Exercise. 

97.  Determine  the   correct   formulas   corresponding  to  the 
following  analyses  and  vapor  densities: 

Carbon.  Hydrogen.  Nitrogen.  Oxygen.      Vapor  Density. 

1.  91.30%  8.70%  45.90 

2.  85.71%  14.29%  35.00 

3-  54-55%  9-°9%  36-36%  66.00 

4.  40.00%  6.67%          53.33  30.00 

5.  92.30%  77°%          26.00 

6.  77-32%  7.63%  15-05%  46.50 


DETERMINATION    OF   ATOMIC   WEIGHTS. 

98.  An  atom  is  the  smallest  particle  of  an  element  that 
is  known  to  form  part  of  a  molecule. 

An  atomic  weight  is  the  ratio  of  the  weight  of  the  smallest 
particle  of  an  element  forming  part  of  a  molecule  to  the  weight 
of  one  atom  of  hydrogen. 

To  determine  the  atomic  weight  of  an  element  it  is  first  nec- 
essary to  analyze  a  series  of  its  compounds  and  determine  the 
percentage  of  the  element  present  in  each. 

The  percentage  of  the  element  present  in  a  single  molecule 
will  of  course  be  the  same  as  that  found  in  the  quantity  of 
material  analyzed.  Therefore  if  the  molecular  weight  of  each 
substance  be  multiplied  by  the  percentage  of  the  element  found 


—128— 

in  it,  the  result  will  be  the  total  weight  of  that  element  in  the 
molecule.  The  smallest  of  these  weights  will  be  the  required 
atomic  weight. 

For  example  it  is  desired  to  find  the  atomic  weight  of  chlorine. 
A  series  of  compounds  of  chlorine  is  analyzed,  and  the  percent- 
age of  that  element  present  in  each  is  determined.  Then  the 
molecular  weights  of  all  the  substances  are  determined. 

The  molecular  weight  of  each  is  multiplied  by  the  percentage 
of  chlorine.  The  result  is  the  number  of  weights  of  chlorine 
in  each  molecule.  An  inspection  of  the  results  shows  at  once 
what  the  atomic  weight  of  the  element  is. 

The  following  table  illustrates  this. 


Substance. 

Vapor 
density. 

Mol. 
weight. 

Per  cent, 
of  chlorine. 

Weight 
present  in 
molecule. 

Hydrochloric  3,cid 

l8.2; 

-36  SO 

07.26$ 

^S.S 

Methyl  chloride 

25.25 

SO  SO 

70.^O 

-JC.C 

Methylene  chloride  

40.50 

00.  OO 

71.72 

71.  0 

Mercuric  chloride  

1  75.  CQ 

271.00 

26.2O 

71.0 

Chloroform  

so.  2=; 

Il8.50 

89.87 

106.5 

Carbon  tetrachloride 

77  OO 

IS4  00 

Q2  21 

142.0 

//.uu 

Since  the  methods  of  obtaining  molecular  weights  are  not 
absolutely  exact,  the  atomic  weights  found  as  above  are  not 
exact.  The  correct  weights  are  found  by  the  analysis  of  com- 
pounds that  admit  of  being  prepared  in  a  condition  of  great 
purity.  See  also  127. 


THE  NATURE  OF  GASEOUS  BODIES. 

99.  All  gaseous  bodies,  that  is,  gases  and  vapors,  are  sup- 
posed to  be  made  up  of  an  almost  infinite  number  of  molecules 
traveling  in  s'traight  lines  with  enormous  velocity.  Their  direc- 
tions are  constantly  altered  by  interference  with  each  other  or 
with  the  walls  of  the  inclosing  vessel,  but  they  are  always 
deflected  in  straight  lines.  The  path  of  a  single  molecule  is 
therefore  made  up  of  a  continuous  broken  line,  the  straight 
portions  of  which  are  very  long  compared  with  the  size  of  the 


I2Q 

molecules.  The  velocity  of  gaseous  molecules  of  a  given 
kind  depends  on  the  temperature  alone.  If  the  temperature 
be  raised,  the  molecular  velocity  is  increased.  The  temperature 
of  a  gas  is  proportional  to  the  average  molecular  velocity. 

The  relative  velocity  of  gaseous  molecules  can  be  determined 
by  measuring  the  relative  rates  at  which  different  gases  escape 
through  very  small  openings  in  very  thin  walls.  From  this  it 
appears  that  the  velocities  are  to  each  other  inversely  as  the 
square  roots  of  the  gas  densities. 

Gaseous  pressure  is  due  to  the  multitude  of  impacts  of  the 
molecules  against  the  walls  of  the  inclosing  vessel. 

100.  Boyle's  Law.     At  constant  temperature  the  volume  of 
a  gas  varies  inversely  as  the  pressure.     That  is,  if  the  volume 
be  reduced  to  one-half  of  what  it  was,  the  pressure  will  be 
doubled. 

This  corresponds  exactly  with  the  above  theory,  for  if  the 
volume  be  decreased  to  one  half  of  what  it  was,  there  will  be 
twice  as  many  molecules  per  unit  of  volume  and  consequently 
twice  as  many  striking  the  walls  and  producing  pressure  as 
there  were  before. 

This  law  may  be  more  generally  expressed  by  the  equation, 

P       V' 

f>=^-  or  PV=P'V 

in  which  P  and  P'  are  the  two  different  pressures  to  which 
a  body  of  gas  is  subjected  at  constant  temperature,  and  V  and 
V  are  the  corresponding  volumes. 

101.  Effect  of  temperature.     If  a  volume  of  gas  at  o°  C. 
be  heated  sufficiently  to  increase  its  temperature  one  degree, 

its  volume  will  be  increased  • of  what  it  was  at  o°  C.     And 

273 

for  each  additional  change  of    one   degree    the    volume  will 

change  -  -  of  what  it  was  at  o°  C.     Therefore  if  the  tempera- 

273 

ture  of  a  volume  of  gas  at  o°  C.  be  decreased  273  degrees,  the 
volume  would  theoretically  become  zero. 

This   temperature,    273    degrees   below    zero,    is   known   as 
absolute   zero.     Temperatures   reckoned   from   this   point  are 
known  as  absolute  temperatures.     Common  temperatures  are 
changed  to  absolute  temperatures  by  adding  273. 
9 


—130— 

102.  The  Law  of  Charles.      At  a  constant  pressure   the 
volume  of  a  body  of  gas  varies  directly  as  its  absolute  tem- 
perature. 

This  law  may  be  more  generally  expressed  by  the  equation, 

JL  =  —  ,  or  VT'  =  V'T,  in  which  V  and  V'  are  the  volumes  of 

a  body  of  gas  at  constant  pressure,  corresponding  to  the  abso- 
lute temperatures  T  and  T'. 

Given  the  Volume  of  a  Body  of  Gas  at  known  Tempera- 
ture and  Pressure,  to  compute  the  Corresponding  Volume, 
Temperature  or  Pressure  under  other  known  Conditions. 

103.  Let  V  be  the  given  volume  at  the  temperature  T  and 
pressure  P.     Assuming  that  the  temperature  remains  constant, 
let  V  be  the  new  volume  corresponding  to  the  new  pressure 
P°.     Then  we  have, 

PV=P°V  or  V;=  ?r. 


Now  let  us  assume  that  the  pressure  remains  constant  and 
that  the  temperature  be  allowed  to  become  T°.  The  gas  will 
assume  a  new  volume,  which  we  will  call  V°.  Applying  the 
law  of  Charles  we  have,  V'T°=V°T  ;  substituting  the  value 
of  V  from  above  we  have, 

PV  _0     -_OT         PV      P°V° 
—  -  x  —  v  T,  or  —  =  —  ^o-. 

This  is  the  general  gas  law.  P,  V  and  T  will  always  be 
known,  also  two  of  the  remaining  factors,  making  it  a  very 
simple  matter  to  determine  the  remaining  one. 

104.  Example.  The  volume  of  a  body  of  gas  is  measured 
at  a  temperature  of  22°  and  a  pressure  of  780  mm.  and  found 
to  be  25  c.c.,  what  would  the  corresponding  volume  be  at  a 
temperature  of  100°  and  a  pressure  1,000  mm.?  Substituting 
in  the  above  formula  we  have, 

25X780  =v  loco  or  v  =  25X780X373 
295  373  295  X  1000 

Note.  Since  the  volumes  of  gases  are  always  measured  sub- 
ject to  atmospheric  pressure,  it  is  customary  to  speak  of  the 
pressures  of  millimeters  of  mercury  ;  the  actual  pressure  being 


that  due  to  a  column  of  mercury  a  certain  number  of  millimeters 
high. 

105.  Normal  Temperature  and  Pressure.  Measured  vol- 
umes of  gases  are  generally  reduced  to  the  corresponding 
volume  at  o°  and  760  mm.  pressure,  known  as  Normal  Tem- 
perature and  Pressure. 

PV       p°y° 


in  which  V°  is  the  required  volume  and  /  the  observed 
temperature,  and  P  the  actual  pressure  to  which  the  gas  is 
subjected  at  the  time  the  volume  V  is  read:  It  is  equal  to  the 
barometric  pressure  minus  the  vapor  pressure  of  the  liquid  in 
contact  with  the  gas,  at  the  temperature  t°. 

Exercise. 

106.  i.  Change  any  given  volume  of  a  gas  from  any  given 
temperature  and  pressure  to  the  corresponding  volume  at  any 
other  temperature  and  pressure. 

2.  Compute  the  temperature  or  the  pressure  of  a  volume  of  a 
body  of  gas  at  known  pressure  and  temperature  corresponding 
to  a  required  volume  and  temperature  or  volume  and  pressure. 

3.  Reduce  a  given  volume  of  gas  from  known  temperature 
and  pressure  to  normal  temperature  and  pressure. 


EVAPORATION  AND  CONDENSATION. 

107.  Evaporation  and  the  Nature  of  a  Vapor.  At  a  con- 
stant temperature,  a  liquid  in  an  inclosed  space  evaporates  until 
the  number  of  molecules  escaping  per  unit  of  time  is  equaled 
by  the  number  of  gaseous  molecules  reentering  the  liquid  con- 
dition. A  definite  condition  of  equilibrium 'is  then  established. 

The  rate  of  evaporation  increases  with  increase  of  tempera- 
ture and  decrease  of  pressure.  The  presence  of  substances 
dissolved  in  the  liquid  not  only  decreases  the  tendency  for  vapor 
to  form,  but  may  cause  the  vapor  to  condense  to  the  liquid  form. 


—132— 

A  vapor  that  is  in  equilibrium  with  its  liquid  is  said  to  be 
saturated.  If  the  pressure  due  to  a  saturated  vapor  in  contact 
with  its  liquid  be  suddenly  reduced,  vapor  will  form  in  the 
liquid  at  other  points  than  on  the  surface,  and  the  liquid  is  said 
to  boil. 

Water  evaporating  at  the  sea  level  in  contact  with  the  atmos- 
phere is  constantly  subject  to  a  pressure  of  about  fifteen  pounds 
to  the  square  inch.  If  the  temperature  be  increased  the  vapor 
pressure  will  finally  become  great  enough  to  lift  the  atmosphere 
away  from  the  surface  and  the  liquid  will  boil.  The  boiling 
point  is  the  temperature  at  which  the  vapor  pressure  equals  the 
atmospheric  pressure.  Water  boils  at  low  temperatures  on  high 
mountains,  and  will  boil  at  ordinary  temperatures  in  a  vacuum. 

A  saturated  vapor  at  a  given  temperature  has  a  definite 
number  of  molecules  per  unit  volume.  These  travel  in  straight 
lines  like  the  molecules  of  a  gas  and  exert  pressure  on  the  walls 
of  the  inclosing  vessel.  If  the  volume  be  decreased  at  constant 
temperature,  some  of  the  vapor  enters  the  liquid  condition,  and 
the  number  of  molecules  per  unit  volume  remains  unchanged. 
In  other  words,  the  volume  of  a  saturated  vapor  may  be 
decreased  without  an  increase  of  pressure. 

1 08.  Condensation    and    Liquefaction.      A    liquid    in    an 
inclosed   space   at   a   given  temperature   is   always   in   contact 
with  its  saturated  vapor.     This  vapor  exerts  a  definite  pressure 
which  is  dependent  on  the  temperature  alone. 

If  the  pressure  be  kept  constant  and  the  temperature  reduced, 
the  vapor  will  constantly  change  to  the  liquid  form,  or  condense. 
If  the  temperature  be  maintained  constant  and  the  pressure 
increased,  condensation  will  again  take  place. 

109.  Critical  Temperature  and  Pressure.     If  a  liquid  be 
contained  in  an  inclosed   space  in   contact  with   its   saturated 
vapor  and  the  temperature  be  increased,  the  liquid  will  evapo- 
rate and  the  pressure  of  the  vapor  will  increase.     This  will 
continue  until  the  density  of  the  vapor  is  the  same  as  that  of 
the  liquid ;  and  the  surface  of  the  liquid,  that  is,  the  line  of 
demarkation  between  the  vapor  and  liquid,  will  disappear,  the 
two  becoming  identical.     The  temperature  at  which  this  takes 
place  is  known  as  the  Critical  Temperature.     In  other  words, 
the  critical  temperature  is  that  above  which  no  amount  of  com- 
pression will  cause  the  gas  to  condense.     Critical  Pressure  is 


—133- 

the  pressure  of  a  vapor  just  sufficient  to  condense  it  at  its 
critical  temperature. 

Effect  of  Substances  in  solution  on  the  Boiling  and  Freezing 

Point. 

no.  The  vapor  pressure  of  a  solution  of  a  substance  is  less 
than  that  of  the  pure  solvent.  This  is  because  the  presence  of 
the  molecules  of  the  dissolved  substance  not  only  tends  to  pre- 
vent the  escape  of  molecules  of  the  solvent  but  actually  attracts 
some  of  those  that  have  escaped,  back  into  the  liquid. 

If  a  solution  of  some  substance  in  water  be  placed  in  the 
same  inclosed  space  with  some  water,  and  the  temperature  be 
constant,  the  water  will  entirely  evaporate  and  condense  in  the 
solution.  This  is  because  the  vapor  pressure  over  the  solution 
can  never  become  as  great  as  the  pressure  over  the  water. 
Therefore  equilibrium  cannot  be  established  until  the  water  has 
entirely  evaporated.  As  the  space  was  inclosed,  this  can  only 
take  place  as  a  result  of  the  condensation  of  the  vapor  into  the 
solution. 

in.  The  Boiling  Point  Law.  If  different  portions  of  the 
same  solvent  contain  the  same  number  of  particles  of  dissolved 
substances  per  unit  volume,  the  boiling  point  will  be  raised  by 
the  same  amount. 

Assuming  for  the  present  that  there  is  no  dissociation  in 
solution,  this  gives  a  very  simple  method  of  getting  molecular 
weights.  It  is  only  necessary  to  find  the  elevation  of  the  boiling 
point  produced  by  a  gram  molecule  of  a  known  substance  in 
a  suitable  solvent,  then  add  to  an  equal  volume  of  the  same 
solvent  sufficient  of  the  unknown  substance  to  produce  the  same 
elevation  of  the  boiling  point.  When  this  has  been  done  one 
gram  molecule  of  the  unknown  substance  must  have  been  used. 

112.  The  Freezing  Point  Law.  Equal  numbers  of  par- 
ticles per  unit  volume  in  different  portions  of  the  same  solvent 
produce  equal  depressions  of  the  freezing  point.  This  again 
gives  a  method  for  determining  molecular  weights.  It  is  only 
necessary  to  compare  the  depression  of  the  freezing  point  pro- 
duced by  a  substance  whose  molecular  weight  is  known  with 
that  produced  by  a  definite  weight  of  a  substance  whose  molecu- 
lar weight  is  desired. 


—134— 
Exercise. 

113.     What  will  be  the  final  condition  of  equilibrium  in  each 
of  the  following  cases?     Give  the  full  explanation. 

1.  A  liquid  in  an  inclosed  space  at  a  constant  temperature. 

2.  Two  portions  of  the  same  liquid  in  an  inclosed  space  at  a 
constant  temperature,  one  portion  being  above  the  other. 

3.  Two  portions  of  the  same  liquid  contained  in  the  same 
inclosed  space,  at  different  constant  temperatures. 

4.  A  water  solution  of  some  substance  in  the  same  inclosed 
space  with  a  vessel  of  water  at  constant  temperature. 


OSMOTIC   PRESSURE. 

114.  Dissolved  substances  exert  a  pressure  in  solution  which 
is  exactly  analogous  to  gaseous  pressure.  This  pressure  is, 
however,  not  appreciable,  unless  the  solution  is  contained  in  a 
vessel  provided  with  a  semi-permeable  membrane,  and  the  whole 
immersed  in  water.  Such  a  membrane  allows  the  passage  of 
the  water,  but  will  not  allow  the  passage  of  molecules  of 
dissolved  substances. 

Suppose  some  sugar  solution  were  placed  in  a  rather  large 
glass  tube,  the  bottom  of  which  had  been  closed  with  a  semi- 
permeable  membrane  and  the  lower  end  placed  in  a  jar  of  water. 
The  solution  would  rise  in  the  tube,  and  water  would  pass  from 
the  jar  through  the  membrane  into  the  tube  until  the  pressure 
due  to  the  column  of  liquid  was  equal  to  the  osmotic  pressure  of 
the  diluted  solution. 

Again,  suppose  one  gram-molecule  of  sugar,  342  g.,  be  placed 
in  a  semi-permeable  cell  of  one  liter  capacity,  the  whole 
immersed  in  water  and  the  osmotic  pressure  measured.  It 
would  be  found  to  be  22.4  atmospheres.  At  a  higher  tempera- 
ture the  pressure  would  become  higher.  If  the  sugar  were  con- 
sidered as  a  gas,  one  liter  of  the  gas  would  weigh  y2  its 
molecular  weight  times  .0896,  which  would  be  15.32  g.  342 
grams  of  sugar  in  the  gaseous  form  at  normal  temperature  and 
pressure  would  therefore  occupy  22.4  liters.  If  this  volume 
were  condensed  to  one  liter  the  pressure  would  be  22.4  atmos- 
pheres. 


—135— 

The  Laws  of  Osmotic  Pressure. 

115.  It  should  be  noted  that  these  laws  are  identical  with  the 
laws  of  Boyle,  Charles  and  Avogadro. 

1.  With  a  fixed  quantity  of  material  in  solution,  at  a  constant 
temperature,    the    osmotic    pressure    varies    inversely    as    the 
volume. 

2.  With  a  fixed  volume  and  quantity  of  material,  the  osmotic 
pressure  is  proportional  to  the  absolute  temperature. 

3.  Equal  volumes  of  the  same  liquid  at  the  same  temperature 
and  showing  equal  osmotic  pressures  contain  equal  numbers  of 
particles. 

Computation  of  Osmotic  Pressures. 

116.  One  gram-molecule  of  any  substance  in  the  gaseous 
form  at  normal  temperature  and  pressure  occupies  22.4  liters. 

The  problem  is  to  find  what  the  pressure  would  be  if  one 
gram-molecule  or  fraction  thereof  of  any  material  is  placed  in 
an  osmotic  cell  at  a  given  temperature. 

The  general  gas  law  may  be  utilized. 

p°v°    py 

T°  T  ' 

P°  is  a  pressure  of  one  atmosphere. 

V°  is  the  volume  of  the  gram-molecule  at  normal  temperature 
and  pressure  equal  to  22.4. 
T°  is  273°  absolute. 
The  expression  then  becomes, 


T          273  * 

P  is  the  osmotic  pressure,  V  the  volume  of  the  cell  and  T  the 
absolute  temperature. 

T      22.4 

-T  —  T  _   /\     ~  . 

V       273 

Example.     What  would  be  the  osmotic  pressure  of  114  grams 
of  sugar  in  one  liter  of  water  at  a  temperature  of  100°  C.  ? 

Substituting-    in    the  formula   we    have,    P  =    —    X   -  -  . 

i  273 

This  would  give  the   pressure   due  to  one  gram-molecule  of 
sugar.     Since  only  114  g.  were  used  the  pressure  would  be 

114 

-  as  great. 
342 


-136- 

The  osmotic  pressures  produced  by  very  dilute  solutions  of 
binary  salts  and  strong  binary  acids  are  found  to  be  nearly  twice 
as  great  as  the  pressure  produced  by  the  corresponding  molec- 
ular quantity  of  sugar.  This  is  due  to  the  fact  that  these 
substances  are  almost  completely  dissociated  in  very  dilute  solu- 
tion. 118.  This  results  in  the  production  of  twice  as  many 
particles  in  the  solution  and  therefore  twice  the  pressure. 

Exercise. 

117.     Compute  the  osmotic  pressures  in  the  following  cases: 

1.  10  g.  of  sugar  in  one  liter  of  water  at  a  temperature  of  o°C. 

2.  10  g.  of  sugar  in  one  liter  of  water  at  a  temperature  of 

273°  c. 

3.  Hydrochloric  acid  is  90%  dissociated  in  a  solution  contain- 
ing .1  gram-molecule  to  the  liter.     What  would  be  the  osmotic 
pressure  produced  by  3.65  g.  of  hydrochloric  acid  in  one  liter  of 
water  at  a  temperature  of  o°  C.  ? 

4..  Assuming  sodium  hydroxide  to  be  90%  dissociated  in  a 
solution  containing  .1  gram-molecule  to  the  liter,  what  would 
the  osmotic  pressure  due  to  a  solution  of  4  g.  in  one  liter  of 
water  at  a  temperature  of  100°  C.  ? 


IONS  AND  IONIZATION. 

118.  Many  substances,  on  being  dissolved  in  water  or.  in 
other  solvents,  are  dissociated,  split  up,  into  at  least  two  parts 
called  ions. 

These  ions  are  the  same  in  many  cases  as  what  we  have 
already  called  radicals,  except  that  they  carry  charges  of  elec- 
tricity. The  positive  radical  carrying  a  positive  charge  is 
known  as  a  positive  ion,  and  the  negative  radical  carrying  a 
negative  charge  as  the  negative  ion. 

If  a  molecule  contains  several  positive  radicals  and  one  or 
more  negative  radicals,  it  splits  up  at  first  into  one  positive  ion, 
consisting  of  one  positive  radical  and  one  complex  negative 
ion.  On  extreme  dilution  it  would  split  up  into  as  many  ions 
as  it  had  radicals. 

An  ion  is  indicated  as  a  radical  with  a  plus  or  minus  sign 
above. 


—137- 

Thus  we  say  sodium  chloride  when  put  in  a  solution  in  water 
gives  positive  sodium  ions  and  negative  chlorine  ions,  repre- 
sented thus : 

NaCl     =  Na+Cl 
Or  H2SO4  =  H+HSO4 

H3P04  =  H+H~P04. 

It  is  understood  that  the  total  positive  charge  of  all  the  posi- 
tive ions  is  exactly  equal  to  the  total  negative  charge  of  all  the 
negative  ions. 

The  Degree  of  Dissociation. 

119.  All  the  strong  acids  and  bases  and  all  salts-  are  almost 
completely  dissociated  in  dilute  solution. 

Substances  that  are  about  90%  dissociated  in  a  solution 
containing  .1  gram-molecule  per  liter: 

Hydrochloric  acid,  HC1. 
Hydrobromic  acid,  HBr. 
Hydroiodic  acid,  HI. 
Chloric  acid,  HC1O3. 
Bromic  acid,  HBrO3. 
Nitric  acid,  HNO3. 
Potassium  hydroxide,  KOH. 
Sodium  hydroxide,  NaOH. 
Barium  hydroxide,  Ba(OH)2. 
Strontium  hydroxide,  Sr(OH)2. 

Some  substances  that  are  .very  slightly  dissociated  in 

solution. 

Water,  H2O. 
Carbonic  acid,  H2CO3. 
Acetic  acid,  HC2H3O2. 
Hydrogen  sulphide,  H2S. 
Hydrocyanic  acid,  HCN. 
Ammonium  hydroxide,  NH4OH. 
Boric  acid,  H3BO3. 
Aluminum  hydroxide. 
Ferric  hydroxide. 
Zinc  hydroxide. 


-138- 

The  Solution  of  Metals  in  Acids. 

1 20.  All  metals  when  placed  in  liquids  have  a  tendency  to 
send  off  ions  into  the  liquid.     These  ions  carry  positive  charges 
of  electricity,  and  the  metal  is  left  with  a  corresponding  negative 
charge.     The  solution  does  not  proceed  to  any  great  extent  in 
water  or  in  liquids  that  have  no  marked  chemical  action  on 
the  metal,  on  account  of  the  attraction  of  the  negatively  charged 
metal  for  the  positively  charged  ions.     The  tendency  of  a  metal 
to  send  off  ions  may  be  called  its  solution  pressure.     This  pres- 
sure varies   greatly   with   the   different   metals,   platinum  and 
gold  having  very  low  solution  pressures,  and  zinc,  magnesium 
and  sodium  very  high  pressures. 

If  a  piece  of  zinc  or  other  metal  with  a  high  solution  pres- 
sure be  placed  in  dilute  sulphuric  acid,  it  at  once  begins  to 
send  off  ions.  The  hydrogen  ions  from  the  solution  at  once 
neutralize  the  negative  charge  on  the  zinc  and  become  atomic 
or  nascent  hydrogen.  These  combine  with  each  other  at  once 
and  are  set  free  as  gaseous  hydrogen.  The  reaction  proceeds 
rapidly  with  extensive  evolution  of  hydrogen  until  the  zinc  is 
all  dissolved,  the  acid  all  used  up  or  the  solution  becomes  so 
saturated  with  zinc  ions  that  their  osmotic  pressure  is  equal  to 
the  solution  pressure  of  the  metal. 

The  Precipitation  of  Metals  from  Solution. 

121.  If  a  metal  having  a  high  solution  pressure  be  placed 
in  a  solution  of  a  salt  of  a  metal  having  a  low  solution  pressure, 
the  latter  will  be  deposited  in  the  metallic  form  on  the  metal 
of  high  solution  pressure.     The  amount  deposited  will  be  the 
exact  chemical  equivalent  of  the  amount  of  the  metal  that  goes 
into  solution. 

Suppose  a  large  piece  of  copper  be  placed  in  a  solution  of 
silver  nitrate.  The  copper,  having  a  higher  solution  pressure 
than  the  silver,  will  send  ions  into  solution ;  the  silver  ions  will 
go  to  the  copper  plate,  where  they  will  give  up  their  positive 
charges  and  appear  as  metallic  silver.  If  sufficient  copper  is 
present,  all  the  silver  will  be  thrown  out  of  solution.  The 
amount  will  be  two  atoms  of  silver  for  each  atom  of  copper 
dissolved. 


The  Simple  Galvanic  Cell. 

122.  If  two  different  metals  be  placed  in  the  same  liquid 
they  each  send  off  ions.     Each  becomes  negatively  charged. 
If  they  be  connected  outside  of  the  liquid  by  a  wire,  a  current 
will  be  found  to  flow  through  the  wire  from  the  one  having 
the  lower  solution  pressure  to  the  one  having  the  higher.     This 
is  because  the  one  having  the  higher  pressure  constantly  sends 
positive  ions  into  the  solution,  and  hydrogen  ions,  or  some  other 
positive  ions  from  the  solution,  are  carried  to  the  other  metal, 
where  they  give  up  their  positive  charges  and  are  set  free. 

The  Daniell  Cell. 

123.  In  this  cell  we  have  a  zinc  plate  surrounded  by  a  solu- 
tion of  zinc  sulphate  in  one  part  of  the  jar  and  a  copper  plate 
surrounded  by  a  saturated  solution  of  copper  sulphate  in  another 
part.     The  two  solutions  are  kept  separate  either  by  the  force 
of  gravity  or  by  a  porous  cup.     When  the  two  plates  are  con- 
nected by  a  wire,  the  zinc,  having  the  greater  solution  pressure, 
sends  off  zinc  ions  into  the  solution,  thereby  forcing  the  copper 
ions  to  the  copper  plate,  where  they  give  up  their  electrical 
charge  and  appear  as  a  deposit  of  metallic  copper. 

PASSAGE    OF    THE    ELECTRIC    CURRENT 
THROUGH    LIQUIDS. 

124.  Electrolyte,  any   substance  which,   in  the  melted  or 
dissolved  condition,  will  allow  the  passage  of  the  electric  current. 

Electrodes,  the  terminals  immersed  in  a  liquid. 

The  positive  electrode  is  the  one  by  means  of  which  the 
current  enters  the  liquid,  called  the  anode. 

The  negative  electrode  is  the  one  by  means  of  which  the 
current  leaves  the  liquid,  called  the  cathode. 

Only  Liquids  that  contain  Ions  will  allow  the  passage  of 

the  Current. 

Pure  water  is  almost  entirely  undissociated,  and  therefore  acts 
like  a  non-conductor.  If,  however,  any  substance  that  is  highly 
dissociated  in  water  solution  be  added,  even  in  very  small  quan- 
tity, the  solution  at  once  becomes  a  good  conductor. 


—140— 

Let  fig.  19  represent  a  vessel  containing  a  dilute  solution  of 
sodium  chloride,  and  let  A  be  the  positive  electrode  and  B  the 
negative  electrode ;  then, 

1.  A  has  a  positive  charge  of  electricity  on  it  and  B  a  negative 
charge. 

2.  When  the  salt  was  dissolved,  it  was  dissociated  into  posi- 
tive sodium  ions  and  negative  chlorine  ions. 

3.  Since  unlike  kinds  of  electricity  attract  each  other,  the 
chlorine  ions,  being  negatively  charged,  move  toward  the  posi- 
tive electrode  and  there  give  up  their  negative  charges  and 
become  atomic  or  nascent  chlorine.     These  soon  unite  with  each 
other,  forming  molecular  or  gaseous  chlorine,  which  is  set  free. 
In  like  manner  the  sodium  ions  go  to  the  negative  electrode 
and  give  up  their  charges ;   but  the  sodium  enters  into  reaction 
with  the  water   forming  sodium  hydroxide,   and  hydrogen  is 
set  free. 


Fig.   19. 

If  the  positive  ion  is  a  metal  that  does  not  react  with  the 
liquid,  it  is  generally  thrown  out  of  solution  in  the  form  of  a 
deposit  on  the  electrode. 

The  negative  ion,  when  it  is  a  single  atom  of  a  non-metal, 
is  generally  set  free;   if  it  is  a  complex  oxygen  ion  it  usually 
reacts  with  the  water  to  form  an  acid,  and  oxygen  is  set  free. 
Thus, 

+  - 

Cu(NO3)2  gives  the  ions  Cu  and  NO.+NO,. 

The  copper,  since  it  does  not  react  with  \vater,  is  disposed 
in  the  metallic  form  as  a  copper  plating  on  the  negative  elec- 
trode, and  the  nitrate  reacts  with  the  water  according  to  the 
equation, 

4NO3  +  2H2O  =  4HNO3  +  O2. 

Or  we  might  assume  that  the  complex  oxygen  ion,  on  losing 
its  negative  charge,  breaks  up  into  the  atomic  or  nascent  con- 


dition.     These  atoms  being  very  active  chemically,  react  with 
each  other  and  with  the  water  to  form  nitric  acid  and  oxygen. 

Exercise. 

Write  all  the  equations  showing  the  changes  involved  in 
the  electrolysis  of  the  following  substances: 

Hydrochloric  acid,  nitric  acid,  sulphuric  acid,  sodium  chlo- 
ride, sodium  nitrate,  sodium  sulphate,  sodium  hydroxide,  cupric 
chloride,  cupric  nitrate,  cupric  sulphate,  silver  nitrate  and  silver 
sulphate. 

125.  Faraday's  Law.  If  any  number  of  electrolytic  cells 
each  containing  different  electrolytes  be  connected  in  series  and 
a  current  passed  through  them  all,  the  actual  quantities  of  the 
various  elements  set  free  will  be  proportional  to  the  correspond- 
ing chemical  equivalents  of  those  elements. 

Let  us  assume  that  the  elements  so  set  free  were  oxygen, 
hydrogen,  copper,  silver,  lead,  zinc,  mercury  and  chlorine,  and 
that  16  weights  of  oxygen  were  formed  in  the  first  cell. 

Then  there .  would  have  been  formed  at  the  same  time,  2 
weights  of  hydrogen,  63.6  of  copper,  216  of  silver,  206.9  °f 
lead,  654  of  zinc,  200  of  mercury  and  71  of  chlorine.  In  other 
words,  the  quantity  of  each  element  set  free  is  proportional 
to  its  atomic  weight  divided  by  its  valence. 


Fig.  20. 

A  simple  Concentration  Cell. 

126.  In  general,  a  current  will  be  produced,  if  two  electrodes 
of  the  same  metal  are  so  arranged  in  a  cell  that  one  is  sur- 
rounded with  a  strong  and  the  other  with  a  weak  solution  of  a 
salt  of  that  metal. 

Let  fig.  20  represent  such  an  arrangement,  the  side  A  con- 
taining a  silver  plate  surrounded  by  a  strong  solution  of  silver 


—1 42— 

nitrate  and  the  side  B  a  silver  plate  surrounded  by  a  weak  solu- 
tion. On  the  side  A,  there  is  a  greater  pressure  of  silver  ions 
against  the  silver  plate  than  there  is  on  the  side  B.  No  great 
quantity  of  silver  can  be  deposited,  however,  on  account  of  the 
inability  of  the  positive  charge  to  leave  the  plate.  If  the  two 
plates  be  connected  by  a  wire,  the  excess  of  positive  charge  from 
the  side  A  is  carried  over  to  the  side  B  and  more  silver  is 
deposited  on  the  plate  A.  The  plate  B  sends  silver  ions  into 
solution.  A  current  will  continue  to  flow  through  the  wire  until 
the  concentration  of  the  silver  nitrate  is  the  same  on  both  sides. 


THERMOCHEMISTRY. 

127.  Law   of   Dulong   and   Petit.      The   product   of    the 
specific  heat  and  the  atomic  weight  of  all  elements  in  the  solid 
state  is  a  constant  and  is  equal  to  about  6.4. 

In  order  to  find  the  approximate  atomic  weight  of  an  ele- 
ment which  exists  in  the  solid  state,  therefore,  it  is  only  neces- 
sary to  divide  6.4  by  its  specific  heat.  For  example,  the 
specific  heat  of  bismuth  is  .0308;  its  atomic  weight  would  be 

^-208.     98. 
.0308 

128.  Heat  of  Formation  and  Heat  of  Decomposition. 

Calorie,  the  amount  of  heat  necessary  to  raise  the  temperature 
of  one  kilogram  of  water  from  o°  C.  to  i°  C.  This  is  known 
as  the  large  calorie.  A  small  calorie  is  one  one-thousandth  of 
a  large  calorie.  In  what  follows,  the  numbers  refer  to  the  large 
calorie. 

Heat  of  formation  is  the  number  of  calories  that  are  set  free 
or  absorbed  when  one  gram-molecule  of  a  substance  is  formed 
from  the  elements. 

Heat  of  decomposition  is  the  number  of  calories  that  are  set 
free  or  absorbed  when  one  gram-molecule  of  a  compound  is 
decomposed  into  its  elements. 

H2  4-   C12  =  2HC1      +    22  Cal. 

'    2K  +     S  =    K2S       +  101  Cal. 

2H2  +    O2  =  2H2O      +    68  Cal. 

H2  +  S  +  2O2  =  H2SO4    +  190  Cal. 

2Na  +  S  +  2O2  =  Na2SO4  +  328  Cal. 


—143— 

I2g.  Heat  of  reaction  is  the  number  of  calories  set  free  or 
absorbed  when  molecular  quantities  of  substances  enter  into 
reaction. 

An  exothermic  reaction  is  one  in  which  the  heat  of  reaction 
is  positive,  or  in  which  heat  is  set  free ;  an  endothermic  reac- 
tion is  one  in  which  the  heat  of  formation  is  negative,  or  in 
which  heat  is  absorbed. 

NaOH  +  HC1  =  NaCl  +  H2O  +  13.7  Cal. 
2N2       +  O2     =  2N2O  —  17.4  Cal. 

The  heat  of  reaction  is  equal  to  the  difference  between  the 
algebraic  sum  of  the  heats  of  formation  of  the  factors  and  the 
heats  of  formation  of  the  products. 

The  heat  of  formation  will  vary  according  to  state  in  which 
the  substance  exists  in  the  reaction.  Table  No.  VII  gives  the 
heats  of  formation  of  some  common  compounds  in  the  various 
states  in  which  they  may  be  formed. 

Exercise. 

130.  Making  use  of  the  values  in  table  VII,  work  out  the 
heats  of  reaction  in  the  following  cases: 

Sodium  oxide,  solid,  with  sulphur  trioxide,  solid. 

Nitrogen  pentoxide,  solid,  with  sodium  oxide,  solid. 

Sodium  hydroxide,  dissolved,  with  hydrochloric  acid, 
dissolved. 

Zinc  with  sulphuric  acid,  dissolved. 

Example.  What  will  be  the  heat  of  reaction  if  potassium  is 
dissolved  in  hydrochloric  acid? 

K  (solid)  +  HC1  (dissolved)  =  KC1  (dissolved)  +  H  (gas). 
39.3  Cal.  ioi.2Cal. 

101.2  —  39.3  =  61.9  Cal. 


—144— 


CHEMICAL    EQUILIBRIUM    AND    THE    LAW    OF 
MASS   ACTION. 


Chemical  Equilibrium. 

131.  Heretofore  in  this  book  all  chemical   reactions  have 
been  represented  by  definite  chemical  equations,  in  which  very 
definite   quantities   of   substances   are   represented   as   entering 
into   chemical   reaction   with   each   other  and   producing   very 
definite    quantities    of   other   substances.      If   the    reaction   be 
imagined  to  take  place  in  an  inclosed  space  of  such  a  nature 
that  none  of  the  products  can  escape,  it  wall  never  go  on  to 
completion  as  indicated  by  the  equation,  but  will  proceed  until 
a  definite  state  of  equilibrium  is  established  between  the  factors 
and  the  products.     The  quantities  of  the  various  substances  in 
the  final   state  will  vary   with  the  temperature,   the   chemical 
affinities  of  the  reacting  substances  and  the  concentration  of 
these  substances  that  were  originally  present. 

In  other  words,  equilibrium  will  be  established  when  the 
velocity  of  the  factors  forming  the  products  is  just  equaled  by 
the  velocity  of  the  products  reforming  the  factors. 

In  this  sense  all  reactions  are  reversible. 

If  A  and  B  enter  into  reaction  to  produce  two  other  sub- 
stances C  and  D,  then  there  will  be  in  the  final  condition  not 
only  some  of  B  and  D  but  also  some  of  A  and  B. 

If  it  is  desired  to  represent  the  possibility  of  a  reversed  reac- 
tion taking  place,  it  is  customary  to  substitute  a  double  arrow 
for  the  equality  sign,  thus, 

A+B  ^=±  C+D. 
Examples:  C+H2O  7"^  CO+H, 

Na3SO4+2HCl  ^=±  2NaCl+H2SO4 

H2S-fZnCl3  ^=±  ZnS+2HCl. 

Completed  Reactions. 

132.  If  any  of  the  products  of  a  reaction  be  allowed  to 
escape  from  the  system,  or  if  any  are  formed  that  are  incapable 
of  producing  a  reversed  reaction,  the  original  action  goes  on 
endeavoring   to    establish    the    required    equilibrium    until    the 
original  reaction  is  virtually  complete. 


—145— 

Example  I.  Precipitates.  If  one  of  the  compounds  formed 
in  solution  is  insoluble,  it  will  separate  at  once  as  a  precipitate, 
the  equilibrium  will  be  destroyed,  and  more  of  the  factors  will 
combine  until  one  of  them  is  entirely  used  up. 

BaCl2  +  K2S04  =  BaS04  +  2KC1. 

CaCl2  +  Na2CO3  =  CaCO3  +  2NaCl. 

Cu(N03)2  +  H2S  =  CuS  +  2HN03. 

Example  2.  One  of  the  products  may  be  volatile  at  the  tem- 
perature at  which  the  reaction  takes  place,  in  which  case  it  will 
escape  in  the  form  of  a  gas  and  allow  the  reaction  to  proceed  to 
completion. 

Na2C03  +  2HC1  =  EuO  +  CO,  +  2NaCl. 

FeS  +  H,S04  =  FeS04  +  H2S. 

KCN  +  HC1  =  KG  +  HCN. 

Na2S03  +  2HC1  =  2NaCl  +  SO2  +  H2O. 

Example  3.  One  of  the  products  may  be  water.  In  this 
case  the  water  may  or  may  not  produce  a  reversed  reaction.  In 
most  of  the  cases  of  neutralization  of  an  acid  by  a  base,  the 
water  formed  cannot  reproduce  the  acid  and  base  by  its  action 
on  the  salt. 

NaOH  +  HC1  =  NaCl  +  HOH. 
H2S04  +  2NaOH  ==  Na2SO4  +  2HOH. 

THE    LAW    OF    MASS   ACTION. 

133.  The  above  examples  are  all  special  cases  illustrating  the 
Law  of  Mass  Action.  According  to  this  law,  when  a  reaction 
has  reached  a  condition  of  equilibrium,  the  ratio  of  the 
products  of  the  concentrations  of  the  substances  on  the  two 
sides  of  the  equation  is  always  equal  to  a  constant,  no  matter 
what  the  concentration  was  at  the  start.  This  is  strictly  true 
only  of  equations  containing  one  molecule  of  a  kind. 

By  concentration  is  meant  the  number  of  molecular  weights 


A  substance,  A,  reacts  with  B  to  form  C  and  D.  Let  a,  b,  c 
and  d  represent  the  resulting  concentrations.  The  equation 
representing  the  reaction  would  be, 

A+B  «=±  C+D 
a     b  c     d. 

10 


—  146  — 
According:  to  the  law  then, 

5*-k 
cd"  k' 

a  constant.    This  constant  will  vary  with  the  chemical  affinities 
involved  and  with  the  temperature. 

If  the  equation  indicated  two  or  more  molecules  of  any  one 
of  the  substances  the  concentration  would  have  to  be  raised  to 
the  corresponding  power.  Thus  if  there  were  two  molecules  of 
A,  the  expression  would  become, 

a'b 

-r  =k. 

cd 

The  law  admits  of  almost  universal  application  to  chemical 
reactions  as  well  as  physical.  The  following  examples  illus- 
trate its  application. 

Example  i.  Solution  of  Gases  in  Liquids. 

134.     Consider  the  case  of  the  solution  of  nitrogen  in  water. 

N2  gas  <    *  Na  dissolved. 

When  a  condition  of  equilibrium  has  been  reached,  the  ratio 
of  the  concentration  of  the  nitrogen  in  the  gaseous  form  to  the 
concentration  in  the  dissolved  form  must  be  a  constant. 

What  would  be  the  effect  of  increasing  the  pressure  on  the 
gas? 

That  is,  in  the  expression 


if  we  increase  a,  what  must  happen  to  b  in  order  that  k  remain 
the  same?  Evidently  b  must  increase  in  proportion.  In  other 
words,  if  the  pressure  of  a  gas  in  contact  with  a  liquid  be 
increased,  more  of  the  gas  will  dissolve  in  the  liquid.  The  fol- 
lowing law  is  a  direct  consequence  of  the  law  df  Mass  Action  : 

Henry's  Law.  At  a  given  temperature  the  amount  of  gas 
that  dissolves  in  a  liquid  is  proportional  to  the  pressure.  This 
does  not  hold  for  those  gases  that  are  extremely  soluble,  such 
as  ammonia,  hydrochloric  acid,  etc. 

Example  2.  A  Liquid  in  equilibrium  with  its  Vapor. 

135.  H2O  vapor  ^Z±  H2O  liquid. 

a  b 


—147— 

Since  the  concentration  of  the  water  cannot  be  changed,  the 
concentration  of  the  vapor  cannot  be  altered.  Any  attempt  to 
increase  it  results  in  transforming  some  of  it  to  the  liquid 
condition. 

Example  3.     A  solid  in  contact  with  its  solution. 

136.  NaCl  dissolved  T~~*  NaCl  solid. 

a  b 

Since  the  concentration  of  the  solid  cannot  be  changed,  b  is 
a  constant,  therefore  a  cannot  be  changed.  Any  attempt  to 
increase  the  concentration  of  a,  by  reducing  the  volume  of  the 
liquid,  results  in  causing  some  of  the  salt  to  crystallize. 

Example  4.     Dissociation  of  Ammonium  Chloride. 

137.  If  ammonium  chloride  be  heated,  it  is  broken  up  into 
ammonia  and  hydrochloric  acid. 

NH4C1  ^=±  NH3+HC1. 
a  be 

When  a  condition  of  equilibrium  is  reached  we  have, 

£-"• 

What  would  be  the  effect  of  having  a  large  quantity  of  ammonia 
present?  Evidently  b  would  be  much  larger  and  c  would  have 
to  become  much  smaller  in  order  that  k  remain  unchanged. 
But  if  there  were  very  little  HC1  formed  there  must  be  very 
little  ammonium  chloride  dissociated.  The  presence  of  a  large 
amount  of  ammonia  prevents  the  dissociation  of  ammonium 
chloride  by  heat. 

Example  5.     lonization. 

138.  When  an  acid,  base  or  salt  is  dissolved  it  is  ionized  to 
a  greater  or  less  extent.     A  definite  equilibrium  is  established 
between  the  undissociated  material  and  the  ions,  and  the  law  of 
Mass  Action  holds  true.     Sodium  chloride  breaks  up  in  solution 
into  sodium  and  chlorine  ions,  and  we  have, 

-f 

NaCl  undissociated  in  sol.   <    >   Cl+Na. 
a  be 


Suppose  a  large  quantity  of  hydrochloric  acid  be  added. 


—148— 

This  acid  is  highly  dissociated  in  solution  and  the  concentra- 
tion of  the  Cl  ions  will  be  increased.  That  means  that  the  con- 
centration of  the  Na  ions  must  be  decreased  or,  what  is  the  same 
thing,  the  concentration  of  the  undissociated  sodium  chloride 
must  be  increased. 

If  the  solution  had  been  saturated,  some  more  salt  would  have 
to  crystallize  out  to  restore  the  equilibrium. 

The  Formation  of  Precipitates. 

139.  In  the  case  of  a  so-called  insoluble  substance  sur- 
rounded by  a  liquid,  we  must  recognize  that  some  of  it  does 
go  into  solution,  and  that  there  is  a  definite  equilibrium  between 
the  solid  and  dissolved  portions.  In  other  words,  when  a  pre- 
cipitate is  formed  the  liquid  is  a  saturated  solution  of  that 
substance. 

The  undissociated  dissolved  substance  is  in  equilibrium  with 
its  ions.  Calling  the  corresponding  concentrations,  a,  b  and  c, 
we  have 

A  =  k. 
be 

Since  a  is  the  concentration  of  a  substance  in  a  saturated  solu- 
tion it  is  a  constant  ;  therefore  the  product  of  the  concentrations 
of  the  ions,  be,  in  such  a  solution  is  a  constant.  The  product 
of  the  concentrations  of  the  ions  of  a  substance  m~^S~Safufate^ 
solution  is  known  as  the  Solubility  Product  of  that  substance. 

If  a  substance  be  added  that  increases  the  concentration  of 
the  ions,  the  product  of  these  concentrations  becomes  greater 
than  the  solubility  product,  and  more  of  the  substance  separates 
as  a  precipitate.  If  a  substance  be  added  that  lessens  the  con- 
centrations of  the  ions,  more  of  the  solid  wrill  go  into  solution. 

For  example,  consider  the  case  of  barium  sulphate. 

-f-f 

BaSO4  undissociated  <    >  Ba+SO4. 
a  be 


Since  this  is  the  case  of  a  saturated  solution,  be  is  the  solubility 
product  and  is  a  constant. 


—149— 

If  an  excess  of  barium  chloride  be  added  to  a  solution  of 
sodium  sulphate,  the  solution  would  for  an  instant  contain  very 
high  concentrations  of  barium  and  sulphate  ions;  the  product 
of  these  concentrations  would  be  much  greater  than  the  solu- 
bility product,  be,  and  barium  sulphate  would  be  precipitated. 

Precipitation  of  a  soluble  Salt  from  Solution  by  the  Addition 
of  an  Excess  of  a  Reagent  having  a  common  Ion. 

140.  Consider  the  case  of  a  saturated  solution  of  sodium 
nitrate. 

The  dissolved  salt,  undissociated,  is  in  equilibrium  with  its 

ions. 

+ 

NaNO3  undissociated  < >  Na+NO8. 
a  be 


and  we  have 


a       , 

i— =  k. 
be 


Since  a  is  constant,  be  is  a  constant  and  equal  to  the  solubility 
product  of  sodium  nitrate. 

Now  suppose  a  large  quantity  of  nitric  acid  be  added  to  the 
solution:  The  concentration  of  the  NO3  ion  will  be  increased, 
and  since  be  is  a  constant,  b  must  decrease.  That  means  that 
more  undissociated  NaNO3  must  be  formed.  The  solution, 
•however,  was  already  saturated,  therefore  salt  will  be  pre- 
cipitated. 

141.  Again,  we  might  consider  the  case  of  a  saturated  solu- 
tion of  silver  acetate, 

+ 

AgCQH3O2  undissociated  < >  Ag+C2H3Oa. 
a  be 

IT— >«• 

be 

a  is  a  saturated  solution  and  therefore  a  constant;  be  must  be 
a  constant.  If  silver  nitrate  be  added  the  concentration  of  the 
silver  ions  will  increase  and  the  concentration  of  the  acetate 
ions  must  decrease.  This  would  cause  a  precipitation  of  some 
of  the  salt. 


—150— 

The  Addition  of  highly  dissociated  Acids  or  Salts  to 
solutions  of  weak  Acids. 

142.  If  a  large  amount  of  sodium  acetate  be  added  to  a 
solution  of  acetic  acid,  the  acid  properties  of  the  solution  almost 
disappear. 

Acetic  acid  is  a  weak  acid;  that  is,  it  is  only  slightly  dis- 
sociated in  solution. 

+ 

HC2H3O2  undissociated  T^  H+C2H3O2. 
a  be 

The  equilibrium  equation  would  be: 


When  the  sodium  acetate  is  added,  the  concentration  of  the 
acetate  ions  becomes  very  much  larger  and  therefore  the  con- 
centration of  the  hydrogen  ions  must  decrease.  As  it  was 
already  very  small,  the  result  is,  almost  no  hydrogen  ions  are 
left  in  the  solution,  and  therefore  the  acid  nature  of  the  solution 
almost  disappears. 

Addition  of  hydrochloric  acid  to  a  solution  of  acetic 
acid  causes  an  almost  complete  suppression  of  the  acetate 
ions. 

+ 

143-       HC,H,O,  undissociated  T^  H+C2H3O2. 
a  be 

The  equilibrium  equation  gives 


Hydrochloric  acid  is  almost  completely  dissociated  in  solu- 
tion ;  its  addition  therefore  causes  a  large  increase  in  the  con- 
centration of  the  hydrogen  ions  with  a  resulting  decrease  in 
the  acetate  ions.  As  the  concentration  of  the  acetate  ions  was 
already  rather  small,  it  is  reduced  to  almost  nothing. 

If  hydrochloric  acid  be  added  to  a  solution  of  hydrogen  sul- 
phide, the  S  ions  are  diminished  to  such  an  extent  that  the 
solution  will  not  blacken  silver  iodide  paper  immersed  in  it. 


Hydrogen  sulphide  is  a  very  weak  acid.     It  is  slightly  dis- 
sociated in  solution  according  to  the  equation, 

+      — 

H2S  undissociated  <    >  2H+S. 
a  be 

The  equilibrium  equation  gives 

—  -k 
b°c~ 

When  the  HC1  is  added,  the  concentration,  b,  is  very  greatly 
increased  and  the  concentration,  c,  correspondingly  decreased. 
This  means  that  as  a  result  there  are  so  few  S  ions  that  there 
is  no  chance  for  the  formation  of  sulphides  with  salts  of  metals. 

The  Necessity  for  an  excess  of  Reagent  to  produce  Complete 

Precipitation. 

144.  In  many  cases  the  solubility  of  the  precipitate  is  suffi- 
ciently appreciable  to  prevent  complete  precipitation  if  exactly 
the  theoretical  amount  of  the  precipitant  is  used. 

In  such  cases  it  is  necessary  to  use  an  excess  of  the  reagent. 
For  example,  consider  the  case  of  the  precipitation  of  calcium 

oxalate.     We  have, 

++ 

CaC2O4  undissociated  <    >  Ca+C2O4. 
a  be 

—  -k 
be" 

a  is  a  constant  in  the  saturated  solution;  be  is  therefore  a 
constant  and  equal  to  the  solubility  product  of  calcium  oxalate. 
Suppose  an  excess  of  ammonium  oxalate  be  added ;  this  sub- 
stance, being  a  salt,  is  highly  dissociated,  the  concentration  of 
the  oxalate  ions  is  greatly  increased,  and  consequently  the  con- 
centration of  the  calcium  ions  must  be  decreased.  This  necessi- 
tates a  precipitation  of  more  calcium  oxalate. 

The  Prevention  of  the  Formation  of  Precipitates. 

145.  Why  cannot  zinc  sulphide  be  precipitated  in  an  acid 
solution  ? 

Assume   that   hydrochloric   acid   is   present   in   considerable 
quantity,  so  that  the  concentration  of  the  H  ions  is  large. 


—1  52— 

Hydrogen  sulphide  is  very  slightly  dissociated. 

+ 

H2S  undissociated  '^~>  2H+S. 
a  be 


The  addition  of  the  hydrochloric  acid  vastly  increases  the  con-, 
centration  of  the  hydrogen  ions  and  consequently  decreases  the 
concentration  of  the  S  ions. 

If  zinc  be  present  in  the  above  solution,  the  product  of  the 
concentration  of  the  zinc  and  sulphide  ions  is  very  small  and 
consequently  less  than  the  solubility  product  of  zinc  sulphide. 
Therefore  zinc  sulphide  will  not  precipitate,  and  if  it  were 
present  it  would  be  dissolved. 

If  copper  were  present  in  the  solution,  the  product  of  the 
concentrations  of  the  sulphur  and  copper  ions  would  be  greater 
than  the  solubility  product  of  copper  sulphide,  and  there  would 
be  a  precipitation  of  the  latter. 

Why  does  Calcium  Oxalate  dissolve  in  Hydrochloric  Acid? 
146.     The  equilibrium  equation  gives 

CaC2O4  undissociated  <    >  Ca+C2O4. 
a  be 


or  since  a  is  a  constant,  be  is  a  constant.  If  hydrochloric  acid 
be  added,  a  large  quantity  of  hydrogen  ions  are  introduced  into 
the  solution,  and  since  oxalic  acid  is  only  slightly  dissociated, 
the  concentration  of  the  oxalate  ions  will  be  decreased  because 
they  form  undissociated  oxalic  acid  and  more  of  the  calcium 
oxalate  must  dissolve  to  make  up  the  loss. 

Action  of  Ammonium  Chloride  on  Magnesium  Hydroxide. 

147.  Ammonium  hydroxide  produces  a  precipitate  of  mag* 
nesium  hydroxide,  because  the  product  of  the  concentrations  o-f 
the  magnesium  ions  and  the  hydroxide  ions  is  greater  than  the~ 
solubility  product  of  magnesium  hydroxide. 


-153- 

Now,  if  a  large  quantity  of  ammonium  ions  be  added  in  the 
form  of  ammonium  chloride,  the  concentration  of  the  hydroxide 
ions  of  the  ammonium  hydroxide  must  decrease.  In  the  equi- 
librium equation, 

+  +      - 

Mg(OH)a   undissociated  ^~*  Mg:+2OH. 
a  be 

a  -k 
b? 

a  is  a  constant,  therefore  be  must  be  a  constant. 

If  the  concentration  of  the  OH  be  decreased,  the  product  be 
would  be  less  than  the  solubility  product  of  magnesium  hydrox- 
ide. That  means,  that  there  would  not  only  be  no  precipitation 
of  magnesium  hydroxide;  but  if  there  were  any  of  the 
undissolved  material  present,  it  would  have  to  go  into  solution 
to  bring  up  the  concentration,  b,  to  the  required  point. 


Hydrolysis. 

148.  Many  salts  are  hydrolyzed  by  water.  That  is  they 
react  with  the  water  to  reform  the  acid  and  base  from  which 
they  were  originally  formed.  This  is  true  of  salts  that  are 
either  compounds  of  strong  acids  with  weak  bases,  strong  bases 
with  weak  acids  or  weak  acids  with  weak  bases. 

By  a  strong  acid  or  base  is  meant  one  that  is  highly  dissociated 
in  water  solution. 

For  example,  a  water  solution  of  potassium  cyanide  reacts 
alkaline.  It  is  a  compound  of  the  strong  base,  KOH,  with  the 
weak  acid,  HCN. 

KCN+HOH  ^=±  KOH+HCN. 

Since  the  KOH  is  highly  dissociated  in  solution,  it  would 
furnish  a  large  quantity  of  OH  ions,  and  since  the  hydrocyanic 
acid  is  a  weak  acid  and  only  slightly  dissociated,  it  would  not 
furnish  enough  hydrogen  ions  to  neutralize  the  alkaline  effect 
of  the  OH. 

Zinc  sulphate  in  water  reacts  acid.  Sulphuric  acid  and  zinc 
hydroxide  are  theoretically  formed.  The  acid  is  highly  disso- 
ciated and  furnishes  hydrogen  ions.  Zinc  hydroxide  is  a  weak 


—154— 

base  and  prevents  the  formation  of  enough  OH  ions  to  neutral- 
ize the  effect  of  the  hydrogen  ions. 

Bismuth  trichloride  is  actually  dissociated  into  bismuth  oxy- 
chloride  and  hydrochloric  acid,  according  to  the  equation, 

BiQ3  +  H2O  =  BiOCl  +  2HC1, 

giving   a   precipitate    of   the   oxy-chloride   and    forming   free 
hydrochloric  acid. 

Exercise. 

Explain  the  reaction  of  the  following  substances  with  water : 
Antimony  chloride ;  zinc  chloride ;  copper  sulphate ;  aluminum 
sulphate ;  potassium  sulphide. 


PART    IV. 
MISCELLANY. 


IMPORTANT  QUESTIONS  RELATING  TO  THE 
FUNDAMENTAL  IDEAS  OF  ELEMENTARY 
CHEMISTRY  WHICH  EVERY  STUDENT 
SHOULD  BE  ABLE  TO  ANSWER  AT  THE 
END  OF  FOUR  MONTHS  OF  STUDY. 

149.     The  numbers  following  the  question  refer  to  the  para- 
graph in  which  the  answer  will  be  found. 

1.  What  is  matter?     1. 

2.  What  is  an  element;   how  many  elements  are  there;   how 
many  of  these  are  common?     2. 

3.  What  is  the  symbol  of  an  element?     3. 

4.  Give  the  symbols  and  names  of  some  common  elements  in 
groups.     4. 

5.  What     is     meant    by     the     expression      "Properties     of 
matter"  ?     5. 

6.  What  is  a  molecule ;  an  atom  ?     6,  7,  8. 

7.  What  is  meant  by  the  chemical  composition  of  a  sub- 
stance?    9. 

8.  What  is  a  physical  change ;  a  chemical  change  ?     Give  five 
examples  of  each.     10,  11. 

9.  What  is  a  chemical  compound?     What  kinds  of  molecules 
would  it  contain?     12. 

10.  What  is  a  mechanical  mixture?     What  kinds  of  mole- 
cules would  it  contain?     13. 

11.  What  is  an  atomic  weight?     14. 

12.  What  is  a  molecular  weight,  and  how  would  that  corre- 
sponding to  a  particular  formula  be  found?     15. 


13.  Give  the  full  meaning  of  the  symbol  of  an  element.     16. 

14.  What  is  the  symbol  or  formula  of  a  compound?     17. 

15.  Give  the  symbols  of  seven  common  compounds.     18. 

1 6.  Give  the  full  meaning  of  the  symbol  or  formula  of  a 
compound  and  illustrate  from  the  formula  of  sulphuric  acid.     19. 

17.  What  is  meant  by  analysis?     20. 

18.  How  may  the  percentage  composition  of  a  substance  be 
worked  out  from  its  formula?    21. 

19.  How  may  the  formula  for  a  substance  be  worked  out 
from  the  analysis?     23. 

20.  What  is  a  binary  compound,  and  how  is  such  a  compound 
named  ?     26. 

21.  What  is  the  law  of  definite  weight?     27. 

22.  What  is  meant  by  the  combining  power  of  an  element  ?    30. 

23.  How  is  it  known  that  elements  have  different  com- 
bining powers?     28-29. 

24.  Give  the  combining  powers  of  the  different  elements  by 
groups.     31. 

25.  If  an  element  has  several  combining  powers,  which  one 
should  ordinarily  be  used?     31. 

26.  How  do  you  tell  what  the  formula  of  a  given  binary 
compound  will  be?     32. 

27.  Into  what  two  general  classes  may  all  the  elements  be 
divided?     25. 

28.  Give  the  symbols  and  names  of  all  the  metals.     25. 

29.  Give  the  symbols  and  names  of  all  the  non-metals.     25. 

30.  Give  the  names  of  all  the  binary  compounds  of  the  metals 
with  the  non-metals.     26. 

31.  Give  the  formulas  and  names  of  all  the  compounds 
of  the  metals  with  the  non-metals.     32. 

32.  When  are  two  elements   said  to  be  chemically  equiva- 
lent?    34. 

33.  How  many  grams  of  any  one  element  would  be  equiva- 
lent to  ten  grams  of  any  other  element?     34. 

34.  What  is  a  reaction?     36.     What  is  a  reagent?     37. 

35.  What  is  a  chemical  equation?     38. 

36.  When  only  have  you  a  right  to  make  use  of  a  chemical 
equation  ?     38. 

37.  Name  some  elements  that  are  known  to  have  two  atoms 
in  the  molecule.     38. 


38.  Write  equations  showing  the  formation  of  binary  com- 
pounds from  the  elements.     39. 

39.  State  the  full  meaning  of  an  equation,  making  use  of 
any  equation  whatever.     41. 

40.  How  many  grams  of  a  binary  compound  may  be  formed 
from  ten  grams  of  one  of  the  elements?     41. 

41.  How  many  grams  of  any  one  element  will  it  take  to  form 
ten  grams  of  a  corresponding  binary  compound?     41. 

42.  What  is  a  ternary  compound?    43. 

43.  What  is  a  radical;  a  positive  radical;  a  negative  radical? 
45,  46,  47. 

44.  Give  the  symbols  and  names  of  all  the  radicals  in 
groups.     50. 

45.  Some  radicals  have  two  combining  powers;    how  is  a 
distinction  made  in  the  name?     51. 

46.  Give  the  names  and  formulas  of  the  compounds  of  all 
the  positive  radicals  with  all  the  negative  radicals.     52,  54. 

47.  What    is   an   acid?      Give   ten    examples,    symbols    and 
names.     50. 

48.  What    is    a    base?      Give   ten    examples,    symbols    and 
names.     50. 

49.  What    is    a    salt?      Give    ten    examples,    symbols    and 
names.     50. 

50.  What  is  an  alkali?     Give  three  examples.     57. 

51.  What  is  a  basic  oxide?     Give  five  examples.     60. 

52.  What  is  an  acidic  oxide?     Give  five  examples.     61. 

53.  What  is  an  alkaline  oxide?     Give  five  examples.     62. 

54.  What  is  an  acid  anhydride?     Give  five  examples.     63. 

55.  State  seven  general  methods  for  the  preparation  of 
salts,  tell  what  els^e  is  formed  and  discuss  each  case.     64,  a, 
b,  c,  d,  e,  f,  g. 

56.  What  is  an  oxidizing  agent?     66. 

57.  What  is  reduction?     67. 

58.  Suppose  a  substance  contains  carbon,  hydrogen,  sulphur 
and  phosphorus;    what  products  will  be  formed  if  it  is  com- 
pletely burned  ?     68. 

59.  Write  equations  showing  what  takes  place  when  some 
of  the  common  elements  are  burned.     67. 

60.  Write    an    equation    showing    the    burning    of    alcohol, 
C2H5OH.     68. 


-158- 

61.  When  an  element  forms  several  different  oxides  how  are 
they  named?     32. 

62.  Mention   three   substances   that   give    up   oxygen   when 
heated.     71. 

63.  Mention  three  substances  that  decompose  water  with  the 
formation  of  oxygen.     71. 

64.  How  can  you  tell  whether  a  compound  containing  oxygen 
and  two  other  elements  is  an  oxidizing  agent  or  not?     71. 

65.  State  the  formulas  of  ten  different  oxidizing  agents  and 
indicate  by  equations  how  they  oxidize.     78. 

66.  Write  an  equation   showing  the  oxidation  of  phos- 
phorus by  nitric  acid.     73. 

67.  What  is  the  action  of  manganese  dioxide  on  hydrogen 
chloride,  hydrogen  bromide  and  hydrogen  iodide?     75. 

68.  Indicate  the  oxidation  of  hydrogen  chloride  with  potas- 
sium permanganate.     76. 

69.  Show  by  an  equation  how  to  change  cuprous  chloride  to 
cupric  chloride  by  means  of  nitric  acid.     77. 

70.  Indicate  the  oxidation  of  ferrous  sulphate  to  ferric  sul- 
phate by  means  of  potassium  permanganate.     77. 

71.  Show  by  equations  the  oxidation  of  each  of  ten  reducing 
agents  by  each  of  ten  oxidizing  agents.     78,  79. 

72.  Mention  ten  good  reducing  agents.     80. 

73.  Show  by  equations  the  reduction  of  ten  metallic  oxides 
with  three  different  reducing  agents.     80,  81,  82. 

74.  What  is  oxidation  ?     65. 

75.  Write  equations  showing  the  formation  of  any  salt  in 
six  different  ways.     64. 

Important  Questions  relating  to  the  more  advanced  Theory. 

150.     i.  State  Avogadro's  Hypothesis.     84. 

2.  How  is  it  known  that  the  hydrogen  molecule  is  divisible? 
86. 

3.  How  is  it  known  that  the  molecule  of  hydrochloric  acid 
contains  equal  numbers  of  hydrogen  and  chlorine  atoms?     87. 

4.  How  is  it  known  that  the  hydrochloric  acid  molecule  con- 
tains but  one  hydrogen  atom  ?     88. 

5.  How  is  it  known  that  the  hydrogen  molecule  contains  two 
atoms?     89. 


—159— 

6.  What  is  gas  density?     90. 

7.  What  is  vapor  density?     91. 

8.  Why   is   the   molecular   weight   equal   to   twice   the   gas 
density  ?     92. 

9.  Describe  two  methods  for  getting  gas  density.     93,  94. 

10.  How  is  a  formula  determined  from  the  analysis  and  the 
gas  density?     95. 

11.  Determine  the  simplest  formula  from  the  analysis.     95. 

12.  How  are  atomic  weights  determined?     98. 

13.  Describe  the  nature  of  gaseous  bodies.     99. 

14.  How  does  the  velocity  of  a  gas  molecule  vary  with  the  gas 
density?     99. 

15.  State  Boyle's  Law.     100. 

1 6.  How  does  the  volume  of  a  body  of  gas  vary  with  the  tem- 
perature?    101. 

17.  State  Charles'  Law.     102. 

1 8.  State  the  general  gas  law.     103. 

19.  Given  the  volume  of  a  body  of  gas  at  definite  tempera- 
ture and  pressure;    find  the  volume,  pressure  or  temperature 
under  other  conditions.     104. 

20.  What  is  meant  by  Normal  Temperature  and  Pressure? 
105. 

21.  Describe  the  nature  of  a  vapor.     107. 

22.  What  is  meant  by  the  Boiling  Point?     107. 

23.  Under  what  conditions  will  a  vapor  condense?     108. 

24.  Define   the   terms    Critical    Temperature    and    Pressure. 
109. 

25.  State  the  Boiling  Point  Law.     111. 

26.  State  the  Freezing  Point  Law.     112. 

27.  What  is  meant  by  osmotic  pressure?     114. 

28.  State  the  laws  of  osmotic  pressure.     115. 

29.  What  would  be  the  osmotic  pressure  of  a  known  weight 
of  a  substance  in  a  given  volume  of  solvent  at  a  given  tempera- 
ture?    116. 

30.  What  are  Ions?     118* 

31.  Mention   ten   substances   that   are   highly   dissociated   in 
solution.     119. 

32.  Mention  ten  substances  that  are  very  slightly  dissociated 
in  solution.     119. 

33.  Explain  the  solution  of  a  metal  in  an  acid.     120. 


— 160 — 

34.  Explain  the  precipitation  of  metals  from  solution.     121. 

35.  Explain  the  action  of  a  simple  galvanic  cell.     122. 

36.  Explain  the  passage    of    the    electric    current    through 
liquids.     124. 

37.  Explain  the  electrolysis  of  potassium  sulphate  solution. 
124. 

38.  State  Faraday's  Law.     125. 

39.  Describe  a  simple  concentration  cell.     126. 

40.  State  the  Law  of  Dulong  and  Petit.     127. 

41.  What  is  meant  by  the  Heat  of  Formation  and  Heat  of 
Decomposition  ?     128. 

42.  What  is  meant  by  the  Heat  of  Reaction?     129. 

43.  What  is  meant  by  an  Exothermic  reaction?     129. 

44.  What  is  meant  by  an  Endothermic  reaction?     129. 

45.  Derive  the  heat  of  reaction  in  a  simple  case,  making  use 
of  Table  VII.     130. 

46.  What  is  meant  by  Chemical  Equilibrium?     131. 

47.  In  what  cases  do  reactions  go  on  to  completion?     132. 

48.  State  the  Law  of  Mass  Action.     133. 

49.  State  Henry's  Law.     134. 

50.  Explain  why  ammonium  chloride  is  not  dissociated  by 
heat  in  the  presence  of  a  large  quantity  of  ammonia.     137. 

51.  What  is  meant  by  the  term  Solubility  Product?     139. 

52.  What  are  the  conditions  necessary  for  the  formation  of  a 
precipitate?     139. 

53.  What  is  the  effect  of  adding  a  highly  dissociated  acid  or 
salt  to  a  solution  of  a  weak  acid?     143. 

54.  Explain  the  prevention  of  the  formation  of  a  precipitate 
of   magnesium   hydroxide   by   means   of   ammonium   chloride. 
147. 

55.  What  is  meant  by  Hydrolysis?     Give  several  examples. 
148. 

College  Examination  Questions. 

1.  Describe  two  modifications  of  oxygen  and  state  how  they 
may  be  prepared. 

2.  Mention  five  substances  that  give  up  oxygen  when  heated. 

3.  Describe   a   test   for   ozone   which   distinguishes   it   from 
bromine  or  chlorine. 


4.  How  may  hydrogen  peroxide  be  prepared  and  describe 
some  of  its  properties? 

5.  How  could  you  recognize  the  products  that  are  formed 
when    sulphur,    carbon,    iron,    phosphorus    and    hydrogen    are 
burned  in  oxygen? 

6.  Describe  your  laboratory  method  for  the  preparation  of 
hydrogen  and  mention  all  precautions  necessary. 

7.  How  may  oxygen  and  nitrogen  be  caused  to  combine? 

8.  How  could  you  prove  that  the  manganese  dioxide  under- 
goes  no   change   when   used   with   potassium   chlorate   in   the 
preparation  of  oxygen? 

9.  What  are  the  constituents  of  the  atmosphere?     Give  the 
approximate  proportions  of  the  four  most  abundant  constituents. 

10.  What  oxides  does  nitrogen  form  and  which  of  these  are 
anhydrides  ? 

11.  State  the  exact  meaning  of  the  symbols,  H;   S;   Fe.    16. 

12.  State    the    exact   meaning    of    the    formulas,    H2 ;     S8 ; 
H2SO4.     19. 

13.  State  the  full  meaning  of  the  equation,  Zn  +  H2SO4  = 
ZnSO4  +  H2.     41. 

14.  State  seven  general  methods  for  the  preparation  of  salts. 
64. 

15.  Write    equations    showing    the    formation    of    sodium 
sulphate  in  six  different  ways.     64. 

1 6.  What  is  a  diabasic  acid;    a  diacid  base;    a  basic  salt; 
an  acid  salt?     Lab.  Ex.  No.  25. 

17.  How  many  liters  of  oxygen  will  13  grams  of  mercuric 
oxide  yield?     152. 

1 8.  How  many  grams  of  sulphuric  acid  would  be  necessary 
to  form  10  liters  of  hydrogen?     152. 

19.  What  are  the  halogens?     Describe  each  briefly. 

20.  What  kind  of  a  chemical  agent  is  required  to  free  chlorine 
from  hydrochloric  acid  ?     75,  76. 

21.  Explain  under  what  conditions,  if  any,  hydrochloric  acid 
acts  upon  the  following  substances  and  what  is  formed  in  each 
case :   Sodium  hydroxide,  zinc  oxide,  zinc,  iron  sulphide,  sodium 
carbonate,    barium    dioxide,    manganese    dioxide.      Write    all 
equations. 

22.  What  is  the  test  for  the  chlorine  radical  ? 

23.  How  could  you  change  sulphur  into  hydrogen  sulphide? 


—162— 

24.  How   did   you   make   sulphuric   acid   and   how   did   you 
recognize  the  product? 

25.  Under  what  condition,  if  any,  does  sulphuric  acid  act 
upon   the    following   substances :     Copper,   lead,   iron,    carbon, 
sulphur,   sodium   carbonate,  barium  dioxide,   calcium  fluoride, 
sodium  chloride  ?     17. 

26.  State  two  general  methods   for  the  preparation  of  the 
halogen  acids. 

27.  Explain  the  use  of  hydrogen  sulphide  in  the  separation  of 
copper  and  zinc. 

28.  How  did  you  make  sodium,  sulphite  from  sulphur  ? 

29.  State  exactly  what  takes  place  when  sodium  hydroxide 
is  neutralized  with  hydrochloric  acid.     118. 

30.  What  is  the  effect  of  mixing  hydrogen  sulphide  water 
and  nitric  acid?     79. 

31.  Describe  the  changes  that  sulphur  undergoes  when  slowly 
heated. 

32.  State   three   different   methods    for   the   preparation    of 
sulphur  dioxide.     71,  65,  64  d. 

33.  How    would    you    change    sodium    sulphate    to    sodium 
chloride?     64  e. 

34.  State  two  methods  for  the  preparation  of  nitrogen. 

35.  Give  the  equation  for  the  action  of  ammonium  hydroxide 
on  hydrochloric  acid.     64  c. 

36.  What  would  be  the  effect  of  heating  ammonium  chloride 
very  hot  ? 

37.  What   is    the    action    of    sulphuric    acid    on    ammonium 
hydroxide?     64  c. 

38.  How  did  you  make  ammonium  nitrate  and  how  did  you 
tell  when  it  was  sufficiently  evaporated?     64  c. 

39.  What    actually    happens    when    ammonium    chloride    is 
heated  ? 

40.  How   may   ammonia  be   obtained    from    an   ammonium 
compound  ? 

41.  What  is  the  effect  of  heating  a  mixture  of  ammonium 
sulphate  and  common  salt? 

42.  How  did   you   make  nitric   acid   and   what   is   its   chief 
characteristic? 

43.  Write  the  equations   showing  the  action  of  nitric  acid 
on  copper. 


-i63- 

44.  If  copper  scale  is  dissolved  in  dilute  nitric  acid,  what 
compound  would  be  formed?     71. 

45.  Suppose  sodium  hydroxide  were  added  to  this  solution, 
and  the  solution  heated,  what  would  be  the  precipitate  be? 

46.  Suppose  copper  scale  were  dissolved  in  dilute  hydrochloric 
acid,  what  compound  would  be  formed? 

47.  If  sodium  hydroxide  were  added  to  this  solution  and  the 
solution  heated,  how  would  this  precipitate  compare  with  the 
last? 

48.  What  is  aqua  regia?     Represent  its  action  by  equations. 

49.  What  is  the  action  of  strong  nitric   acid   on   zinc? 

50.  What  is  the  action  of  very   dilute  nitric  acid   on  zinc? 
Give  equations. 

51.  What  is  the  action  of  an  excess  of  strong  nitric  acid  on 
yellow  ammonium  sulphide?     71. 

52.  How   may   red   phosphorus   be   changed  to   yellow   and 
yellow  to  red? 

53.  What   is    the    effect   of    adding   bromine    water   to    red 
phosphorus?     71. 

54.  What    is    the    effect    of    adding    iodine    water    to    red 
phosphorus?     71. 

55.  What  is  the  effect  of  dissolving  red  phosphorus  in  nitric 
acid,  adding  silver  nitrate  and  ammonia  so  as  not  to  mix? 

56.  Describe  the  effect  of  heating  arsenic  in  closed  tube. 

57.  Describe  the  effect  of  heating  arsenic  in  air. 

58.  What  is  the  effect  of  heating  arsenious  oxide  with  char- 
coal ?     80. 

59.  What  is  the  effect  of  heating  arsenious  oxide  in  closed 
tube? 

60.  What  effect  has  bromine  water  on  metallic  arsenic?     71. 

61.  What  is  the  color  of  silver  arsenater 

62.  What  is  formed  when  arsenic  is  dissolved  in  concentrated 
nitric  acid?     73. 

63.  What  is  the  reaction  between  ammonium  hydroxide  and 
arsenic  trioxide?     What  is  the  color  of  silver  arsenite?     64  g. 

64.  Which  is  the  most  soluble,  arsenious  oxide  or  arsenic 
oxide  ? 

65.  What  is  formed  when  bromine  water  acts  on  arsenious 
oxide?     71. 

66.  What    is    the    action    of    concentrated    nitric    acid    on 
powdered  antimony?     73. 


— 164— 

67.  What  is  formed  when  bismuth  carbonate  is  dissolved  in 
hydrochloric  acid?     64  d. 

68.  When  this  solution  is  poured  into  an  excess  of  water, 
what  is  the  precipitate  ?     148. 

69.  What  are  the  three  allotropic  forms  of  carbon  and  what 
are  your  reasons  for  believing  them  to  be  the  same  substance? 

70.  What  is  the  effect  of  heating  copper  scale  with  charcoal? 
80. 

71.  What  is  the  effect  of  passing  illuminating  gas  over  hot 
cupric  oxide?     80. 

72.  What  effects  are  produced  by  passing  an  excess  of  carbon 
dioxide  into  lime  water?     64  g. 

73.  What  is  the  effect  of  heating  hydrogen  sodium  carbonate? 

74.  What  acid  does  carbon  dioxide  form  with  water?     69. 

75.  Explain  the  action  of  a  water  solution  of  sodium  car- 
bonate on  litmus.     148. 

76.  What  is  common  baking  powder?     Lab.  Ex.  No.  25. 

77.  What  is  the   effect   of  heating  Rochelle   Salts  until   all 
carbon  is  burned? 

78.  How  did  you  make  orthosilicic  acid  ?     Lab.  Ex.  No.  26,  3. 

79.  Mention  three  ways  of  dissolving  sand. 

80.  How  did  you  make  stannous  chloride?     64  a. 

81.  How  would  you  change  stannous  chloride  into  stannic 
chloride?     77. 

82.  What  is  the  action  of  stannous  chloride  upon  mercuric 
chloride?     Lab.  Ex.  No.  27,  2. 

83.  What  is  the  action  of  hydrogen  sulphide  upon  stannous 
chloride;    upon  stannic  chloride? 

84.  How  do  the  common  acids  affect  lead  ? 

85.  What  is  the  action  of  a  soluble  sulphate  upon  a  solution 
of  a  lead  salt? 

86.  WTiat  is  the  effect  of  trying  to  dissolve  red  lead  in  dilute 
nitric  acid?     What  is  supposed  to  be  the  composition  of  red 
lead? 

87.  WThat   is   the    Solvay   or   Ammonia   process   of   making 
sodium  carbonate? 

88.  How  could  you  distinguish  between  carbonate  and  bicar- 
bonate of  soda? 

89.  Describe  how  you  made  sodium  hydroxide  in  the  labora- 
tory.    Lab.  Ex.  No.  29. 


-i65- 

90.  What  is  the  effect  of  adding  concentrated  hydrochloric  to 
a  saturated  solution  of  common  salt?     140. 

91.  What    effect    has    silver    nitrate    upon    a    solution    of 
potassium  chlorate? 

92.  Describe  methods  for  making  the  following  salts :  Sodium 
sulphate  from  sodium  chloride,  sodium  chloride   from  sodium 
sulphate,  sodium  acetate  from  sodium  chloride,  sodium  chloride 
from  hydrogen  di-sodium  phosphate. 

93.  What  oxides  does  copper  form?     Describe  each. 

94.  Describe  the  action  of  the  common  acids  on  copper,  both 
hot  and  cold.     Give  equations  where  there  is  any  action. 

95.  How    could    you    change    cupric    chloride    to    cuprous 
chloride  ?     82. 

96.  How    could    you    change    cuprous    chloride    to    cupric 
chloride?     73. 

97.  What  is  the  effect  of  adding  potassium  iodide  to  copper 
sulphate  solution? 

98.  Describe  all  changes  when  sodium  hydroxide  is  added  to 
a  solution  of  cupric  sulphate  and  the  mixture  heated  to  boiling. 

99.  State   two   methods    for   making   cuprous    oxide.     Lab. 
Ex.  No.  31,  3,  5. 

100.  What  is  the  effect  of  adding  a  piece  of  iron  wire  to  a 
copper  sulphate  solution?     121. 

101.  How  did  you  obtain  pure  silver  from  a  silver  coin? 

1 02.  How  could  you  obtain  silver  nitrate  from  silver  chloride? 

103.  How  can  the  precipitation  of  magnesium  hydroxide  by 
ammonia  be  prevented  without  making  the  solution  acid  ?     147. 

104.  Describe  lime,  slaked  lime,  lime  water,  calcium  hydrox- 
ide, air-slaked  lime. 

105.  How  could  you  distinguish  air-slaked  lime   from  dry- 
slaked  lime? 

1 06.  What  is  the   difference  between  chloride  of  lime   and 
calcium  chloride?     State  the  uses  of  each. 

107.  What  is  the  formula  of  gypsum?     What  is  plaster  of 
Paris  ? 

108.  Why  does  plaster  of  Paris  set? 

109.  Describe   barium   hydroxide.      What   happens   when   a 
solution  of  barium  hydroxide  is  left  exposed  to  the  air?     46  g. 

no.  What   is   the   action   of   zinc   on   potassium   hydroxide 
solution  ? 


— 166— 

in.  What  is  formed  when  potassium  hydroxide  is  added  to 
a  zinc  sulphate  solution  ?     Lab.  Ex.  No.  30,  7. 

112.  What  is  the  action  of  zinc  chloride  on  litmus  and  explain 
the  result?     148. 

113.  What   is    the    action    of    cold    and    hot   nitric    acid    on 
mercury  ? 

114.  What  is  the  effect  of  adding  a  sodium  hydroxide  solu- 
tion to  a  solution  of -mercuric  nitrate? 

115.  What  is  the  effect  of   adding  a  soluble  chloride  to   a 
mercurous  solution  ? 

116.  What  is  the  effect  of  adding  sodium  hydroxide  to  a 
solution  of  a  mercuric  salt? 

117.  What  is  the  action  of  the  common  acids  on  aluminum? 

118.  What  is  the  residue   after  heating  aluminum   chloride 
very  hot? 

119.  What  peculiar  property  has  aluminum  amalgam? 

120.  Describe  chromic  hydroxide. 

121.  How  could  you  change  chromic  sulphate  to  a  chromate? 

122.  How  could  you  change  chromium  in  potassium  dichro- 
mate  to  chromic  sulphate  ? 

123.  What   would  be   the   effect   of   chromic   anhydride   on 
hydrochloric  acid? 

124.  What  is  the  effect  of  adding  oxalic  acid  to  potassium 
permanganate  solution  in  the  presence  of  sulphuric  acid  ?     76. 

125.  Describe  the  effects  of  the  common  acids  on  iron. 

126.  Describe  ferrous  hydroxide. 

127.  How  did  you  change  ferrous  sulphate  to  ferric  sulphate? 
77. 

128.  Describe  ferric  hydroxide. 

129.  How    would    you    change    ferric    hydroxide    to    ferric 
chloride  and  this  to  ferrous  chloride?     64  c,  82. 

130.  Describe  the  effect  of  adding  hydrogen  sulphide  to  both 
ferrous  and  ferric  salts  in  solution.     81. 

131.  How  did  you  make  ferrous  ammonium  sulphate? 

132.  How  could  you  change  potassium  ferrocyanide  to  ferric 
chloride? 

133.  Why  does  not  ammonium  hydroxide  give  a  precipitate 
of  ferric  hydroxide  in  a  solution  of  potassium  ferrocyanide? 

134.  How  could  you  make  chloroform? 

135.  How  could  you  make  ethyl  acetate? 


-i67- 

136.  What  is  the  general  formula  of  an  alum,  and  how  could 
you  make  ammonium  aluminum  alum? 

137.  Explain  why  the  same  amounts  of  heat  are  generated 
when  equivalent  amounts  of  different  acids  are  neutralized  with 
bases. 

138.  Describe    the    properties    and    preparation    of    the    two 
chlorides  of  mercury. 

139.  What  action,  if  any,  has  dry  ammonia  gas  on  litmus 
paper? 

140.  How  is  the  aluminum  ion  detected  ? 

141.  Describe  several  different  tests  for  ferrous  and  ferric 
ions. 

142.  What  acids  does  phosphorus  form?     69. 

143.  How  could  you  make  lead  chloride  and  what  remarkable 
property  has  this  salt? 

144.  How  much  methyl  alcohol  should  be  dissolved  in  one  liter 
of   water   to   give   the   same   pressure  that   5    grams   of   ethyl 
alcohol  give  when  dissolved  in  the  same  volume  of  the  same 
solvent?     115. 

145.  What  is  meant  by  the  term    "weak  acid"    or    "weak 
base"? 

146.  What  is  the  method  for  finding  the  atomic  weight  of  an 
element  like  calcium  that  does  not  form  gasifiable  compounds? 
127. 

147.  State  two  methods  for  making  aluminum  chloride  from 
alum. 

148.  Explain  the  fact  that  the  gas  density  of  sulphuric  acid  is 

24-5- 

149.  Describe  and  explain  what  takes  place  when  a  piece  of 

zinc  is  placed  in  a  solution  of  a  lead  salt.     121. 

150.  Examples  involving  weights  only.     151. 

151.  Examples  involving  weights  and  volumes  of  gases.  152. 

152.  Examples  involving  volumes  of  gases  only.     154. 

153.  Examples  involving  volumes  of  liquids  having  a  certain 
specific  gravity  and  a  certain  per  cent,  of  material  in  solution. 
155. 

154.  Derivation  of  formulas  from  analysis  and  vapor  density. 
95. 

155.  General  Theory.     See  questions  with  references.     150. 


— 168— 


SIMPLE  PROBLEMS   IN   ELEMENTARY 
INORGANIC   CHEMISTRY. 

151.  How  many  grams  of  a  substance  are  necessary  to 
combine   with   or  produce  a   certain  number  of  grams   of 
another  substance? 

Example.  Ten  grams  of  sodium  nitrate  could  be  formed 
from  how  many  grams  of  nitric  acid? 

It  is  generally  best  to  write  an  equation  showing  the  possible 
reaction,  although  if  the  chemical  equivalents  are  surely  known 
this  is  not  necessary. 

?  g.  10  g. 

NaOH  +  HNO3  =  NaNO3  +  H2O. 
63  85    ' 

From  this  it  appears  that  one  molecule  of  nitric  acid  is  exactly 
equivalent  to  one  molecule  of  sodium  nitrate. 
That  is,  that  63  weights  of  nitric  acid  will  exactly  form  85 

weights  of  the  salt.     One  weight  would  form  —  of  85  weights, 
nd  therefore  as  many  weights  would  be  necessary  to  form  ten 

as  >    is  contained  in  10  or  —  X  10. 
63  85 

Or  it  might  be  done  by  proportion,  which  amounts  to  exactly 
the  same  thing.  The  unknown  weight  in  grams  is  to  the 
given  weight  in  grams  as  the  molecular  weight  of  the 
unknown  substance  is  to  the  molecular  weight  of  the  known 
substance. 

This  would  give  -—  =  5-  or  X=  ~-   X  10. 
10       85  85 

152.  Examples  involving  a  weight  of  a  substance  and  the 
volume  of  a  gas. 

Example.  How  many  liters  of  carbon  dioxide,  at  normal 
temperature  and  pressure,  could  be  formed  from  ten  grams  of 
sulphuric  acid? 

10  g. 

Na2CO3  +  H2SO4  =  Na2SO4  +  H2O  +  CO2. 
98  44 


— 169 — 

One  molecule  of  sulphuric  acid  is  equivalent  to  one  molecule 
of  carbon  dioxide.     Ten  grams  of  the  acid  would  therefore 

form    -  X  10  grams  of  carbon  dioxide. 
90 

The  weight  of  one  liter  of  any  kind  of  gas  at  normal 
temperature  and  pressure  is  equal  to  one  half  its  molecular 
weight  multiplied  by  .09. 

Therefore  the  weight  of  a  liter  of  carbon  dioxide  equals 
22  X  .09  =  i  .98  gr.  There  will  be  as  many  liters  in  the  above 

weight  of  the  gas  as  1.98  g.  is  contained  in  -  ^X  10  or  -- 

90  90  X  i  .98 

liters  at  norm.  temp,  and  pressure. 

Example.  How  many  grams  of  sodium  sulphite  would  ten 
liters  of  sulphur  dioxide  form? 

One  liter  of  sulphur  dioxide  weighs  32  X  .09  grams  or  2.88 
g.  Ten  liters  would  weigh  28.8  g. 

SO2  +  2NaOH  =  Na2SO3  +  H2O. 
64  126 

28.8  g.  ?  g. 

One  molecule  of  the  gas  is  equivalent  to  one  molecule  of  the 

126 
salt.     28.8  g.  would  therefore  make  28.8  X  -^—  g.   of  sodium 

sulphite. 

153.  Given  the  volume  of  a  body  of  gas  at  a  certain  temp, 
and  pressure  to  find  the  volume  corresponding  to  any  other 
temp,  and  pressure. 

Example.  A  volume  of  nitrogen. is  measured  at  a  temp,  of 
10°  C.  and  pressure  of  780  mm.  and  found  to  be  10  c.c.  What 
would  be  the  corresponding  volume  at  20°  and  790  mm.  ? 

The  volume  of  a  body  of  gas  varies  as  its  absolute  tem- 
perature. In  this  case  the  temperature  has  increased,  therefore 
the  volume  would  be  increased.  The  absolute  temperatures  con- 
cerned are  283°  and  293°,  therefore  the  given  volume  should 

2QO 

be  multiplied  by  — ~.     This  will  give  the  new  volume,  con- 
sidering temperature  alone. 

The  volume  of  a  body  of  gas  varies  inversely  as  the 
pressure.  The  pressure  in  this  case  has  increased,  therefore 


—i7o— 

the  volume  would  be  decreased  on  account  of  the  pressure 
change.  Therefore  the  above  new  volume  should  be  multi- 
plied by  . 

790 

The  final  reduced  volume  would  be  10  X  — -~  X c.c. 

283       790 

154.     Examples  involving  volumes  of  gases  only. 

Example.  What  volume  of  oxygen  would  be  necessary  to 
exactly  burn  ten  liters  of  alcohol  vapor?  What  would  be  the 
volumes  of  the  gaseous  products? 

C2H60  +  302  =  3H20  +  2C02. 

First  write  a  molecular  equation  representing  the  reaction. 

Equal  volumes  of  all  gases  at  the  same  temp,  and  pressure 
contain  equal  numbers  of  molecules. 

In  other  words,  all  gaseous  molecules  occupy  equal  space. 

The  coefficients  in  a  molecular  equation  represent  at  once  the 
relative  combining  volumes  of  the  indicated  substances  in  the 
gaseous  form. 

That  means  in  the  above  equation  that  one  volume  of  alcohol 
vapor  would  combine  with  three  volumes  of  oxygen  and  form 
three  volumes  of  water  vapor  and  two  volumes  of  carbon 
dioxide.  Or  that  ten  liters  of  alcohol  vapor  would  require 
thirty  liters  of  oxygen  to  burn  it,  and  there  would  be  formed 
thirty  liters  of  water  vapor  and  twenty  liters  of  carbon  dioxide. 


Example  involving  the  per  cent,  of  substance  in  solution 
and  the  specific  gravity  of  the  solution. 

155.  In  the  majority  of  chemical  operations  the  reagents  are 
handled  in  solution,  the  specific  gravity  of  the  liquid  and  the 
per  cent,  of  the  substance  in  solution  being  known. 

The  problem  is  exactly  like  the  simple  case  in  151  except 
that  in  place  of  a  definite  weight  being  given  or  required,  a 
volume  of  a  certain  solution,  containing  a  certain  per  cent,  of 
material  and  having  a  given  specific  gravity,  is  given  or 
required. 

If  a  volume  of  a  liquid  is  given,  find  at  once  the  amount  of 
available  material  it  contains  and  proceed  as  in  151. 


—I/I— 

If  the  volume  of  a  liquid  is  required,  first  find  the  number  of 
grams  required  as  in  151,  and  then  reduce  this  number  to  the 
volume  of  the  solution  necessary  to  give  it. 

It  may  be  that  the  volume  of  one  liquid  is  given  to  find  the 
volume  of  another  liquid  concerned. 

156.  Example  i.  100  c.c.  of  sulphuric  acid  solution  having 
a  s.g.  of  1.84  and  containing  98%  acid,  will  form  how  many 
grams  of  hydrochloric  acid? 

In  all  of  these  examples  if  we  call  S,  the  specific  gravity  of 
the  liquid,  P,  the  per  cent,  of  substance  in  solution,  and  W,  the 
weight  of  material  in  volume,  V,  the  following  expression  holds 
true, 

W  W  W 

SVP  =  Wor  V  =          P=r 


In  using  these  expressions  never  confuse  volumes  of  gases 
with  volumes  of  liquid. 

100  c.c.  of  the  acid  would  contain  100  X  1.84  X  .98  grams 
of  H2SO4,  which  equals  180.3  g. 

1  H2SO4  +  2NaCl  =  Na2SO4  +  2HC1. 

98  73 

180.3  g-  ?  g- 

One  molecule  of  sulphuric  acid  is  equivalent  to  two  molecules 
of  hydrochloric  acid.  .  That  is,  180.3  §"•  °f  sulphuric  acid  would 

produce  —  -^r  X  180.3  £•  of  hydrochloric  acid. 

157.  Example  2.  How  many  c.c.  of  a  barium  chloride  solu- 
tion, having  a  s.g.  of  i.i  and  containing  10%  barium  chloride, 
would  precipitate  2  g.  of  sulphuric  acid  ? 

BaCl2  +  H,SO4  =  BaSO4  +  2HC1. 
208  9§ 

?  g-          2  g. 

—  —•  X  2  gf.  of  barium  chloride   would  be  necessary.     This 

amount  would  be  contained  in  as  many  c.c's.  of  the  solution 

,,,      208  vx          208X2 
as,  SP  =  .11  is  contained  m  W  =  —^-  X2  or  ^~^  -  c.c. 


—172— 

158.  Example  3.  How  many  c.c.  of  a  solution  of  potassium 
permanganate,  one  c.c.  of  which  contains  .1  g.  of  the  salt,  would 
be  necessary  to  oxidize  all  the  iron  contained  in  100  c.c.  of  a 
solution  of  ferrous  sulphate  having  a  s.g.  of  i.i  and  containing 
10%  of  FeSO4? 

SVP  =  W.  i.i  X  ioo  X  .10=  ii  grams  of  FeSO4  con- 
tained in  the  solution  to  be  oxidized. 

ioFeSO4  +  2KMnO4  +  8H2SO4  = 

5Fe2(S04)3  +  K2S04  +  2MnS04  +  8H2O. 
1520  316 

ii  g-  ?  g- 


The  number  of   grams  of   potassium  permanganate  necessary 

ined  in  as 
316X11 


would  be   Xu,  and  this  would  be  contained  in  as  many 

1520 


c.c.  of  the  solution  as  .1  is  contained  in  it,  or 

I520X.I 

159.  Example  4.  How    many    c.c.    of    hydrochloric    acid 
solution  having  a  s.g.  of  1.2  and  containing  40%  acid  would  be 
necessary  to  form  ioo  liters  of  carbon  dioxide?  t 

One  liter  of    carbon   dioxide    weighs  -  L  X   .09  =   1.98  g. 
ioo  liters  would  weigh  198  g. 

CaCO3  +  2HC1  =  CaCl2  +  H2O  +  CO2. 

73  44 

?  g.  198  g. 

*7\ 

The  number  of  grams  of  hydrochloric  acid  would  be  —  X  198  g. 

44 

V=  ^=r  =      7>;  -  =  number  of  cubic  centimeters  of  the 

SP       1.2X40X44 

solution. 

Review  Exercise. 

1 60.  There  are  many  important  chemical  transformations 
in  elementary  chemistry  with  which  the  student  should  be  very 
familiar.     Some  of  these  are  listed  in  Table  I.     The  student 
should  learn  to  state  the  various  steps  as  actually  carried  out  in 
practice,  as  well  as  to  represent  the  successive  changes  by  equa- 
tions.    It  is  safe  to  say  that  any  student  who  can  do  this  as 


—173— 

called  for  in  Table  I.  is  necessarily  well  grounded  in  the  prin- 
ciples of  elementary  chemistry.  If  there  are  two  or  more 
methods  he  should  learn  them  all,  especially  the  common  tech- 
nical method.  If  the  laboratory  or  analytical  method  is  differ- 
ent from  the  technical  method,  that  should  also  be  thoroughly 
learned. 

For  example,  consider  No.  23,  Table  I.  How  change  mer- 
cury to  mercuric  chloride? 

Technical.  Dissolve  the  mercury  in  sulphuric  acid,  getting 
mercuric  sulphate.  Heat  the  mercuric  sulphate  with  sodium 
chloride  and  mercuric  chloride  will  sublime  out. 

Hg  +  2H2SO4  =  HgSO4  +  SO2  +  2H2O. 
HgSO4  +  2NaCl     =  HgCl2    +  Na2SO4. 


Laboratory  method.     Dissolve  the  mercury  in  aqua  regia. 


2HN03       =    H00 
30  +  6HC1  =  3H20  +  3C12. 
Q2    =  3HgCl2 


Adding  these  equations  we  have, 

3Hg  +  2HN03  +  6HC1  =  3HgCl2  +  4H2O 

The  student  is  not  expected  to  invent  any  of  the  reactions, 
he  is  expected  to  look  up  the  actual  methods  in  the  text  book. 
All  reactions  represented  must  be  possible. 


—174— 

TABLE  I. 
REVIEW  EXERCISE. 

161.     Indicate  by  equations  the  following  transformations. 

1.  Sodium  to  sodium  chloride.     64  a. 

2.  Calcium  oxide  to  calcium  nitrate.     64  b. 

3.  Aluminum  sulphate  to  aluminum  hydroxide.    64  e. 

4.  Sodium  chloride  to  sodium  sulphate.     64  d. 

5.  Aluminum  hydroxide  to  aluminum  sulphate.     64  c. 

6.  Calcium  carbonate  to  calcium  chloride.     64  d. 

7.  Sodium  sulphate  to  sodium  chloride.     64  e. 

8.  Sulphur  trioxide  to  sodium  sulphate.     64  f,  g. 

9.  Sodium  chloride  to  sodium. 

10.  Calcium  to  calcium  nitrate.     78,  8. 

11.  Calcium  chloride  to  calcium  nitrate.     64  e,  64  d. 

12.  Barium  nitrate  to  barium  chloride. 

13.  Cupric  sulphate  to  cupric  oxide. 

14.  Mercuric  oxide  to  mercury. 

15.  Mercuric  chloride  to  mercuric  oxide. 

1 6.  Barium  sulphate  to  barium  chloride.     163. 

17.  Strontium  chloride  to  strontium  oxide. 

18.  Silver  to  silver  chloride.     78,  8;   64  e. 

19.  Copper  to '  cupric  sulphate.     71. 

20.  Copper  to  cupric  chloride.     160. 

21.  Cupric  chloride  to  cuprous  chloride.     82, 

22.  Silver  chloride  to  silver  nitrate.     121,  78,  8. 

23.  Mercury  to  mercuric  chloride.     160. 

24.  Mercury  to  mercurous  chloride.     160,  82. 

25.  Stannous  chloride  to  stannic  chloride.     77. 

26.  Ferrous  chloride  to  ferric  chloride.     77. 

27.  Cuprous  chloride  to  cupric  chloride.     77. 

28.  Ferrous  sulphate  to  ferric  sulphate.     77. 

29.  Ferrous  chloride  to  ferric  sulphate. 

30.  Mercurous  chloride  to  mercuric  sulphate. 

31.  Stannous  chloride  to  stannic  sulphate. 

32.  Sulphur  to  hydrogen  sulphide. 

33.  Sulphuric  acid  to  sulphur  dioxide.     71. 

34.  Nitric  acid  to  ammonia. 

35.  Sodium  hydroxide  to  sodium  aluminate. 

36.  Chromous  oxide  to  chromic  acid. 

37.  Potassium  dichromate  to  potassium  chromate. 

38.  Iodine  to  iodic  acid. 

39.  Potassium  dichromate  to  chromic  chloride. 

40.  Manganese  dioxide  to  permanganic  acid. 


—175— 


TABLE   II. 

TABLE  OF  NATURAL  ELEMENTS  AND  COM- 
POUNDS FROM  WHICH  NEARLY  ALL  OTHER 
COMMON  COMPOUNDS  AND  ELEMENTS  MAY 
BE  MADE. 


162.     First  Group. 

Sodium  chloride. 
Sodium  nitrate. 
Potassium  chloride. 

Second  Group. 

Magnesium  carbonate. 
Magnesium  sulphate. 
Calcium  carbonate. 
Calcium  sulphate. 
Strontium  carbonate. 
Strontium  sulphate. 
Barium  carbonate. 
Barium  sulphate. 
Zinc  carbonate. 
Zinc  sulphide. 
Cadmium  sulphide. 
Mercury. 
Mercury  sulphide. 

Third  Group. 

Aluminum  sulphate. 
Aluminum  silicate. 
Borax,  Na2B4O7. 

Fourth  Group. 

Carbon. 

Carbon  dioxide. 
Calcium  carbonate. 
Silicon  dioxide. 
Aluminum  silicate. 
Tin  dioxide. 
Lead  sulphide. 


Fifth  Group. 

Nitrogen. 
Sodium  nitrate. 
Calcium  phosphate. 
Arsenic. 

Arsenic  sulphide. 
Antimony  sulphide. 
Bismuth  sulphide. 

Sixth  Group. 
Oxygen. 

Hydrogen  oxide. 
Sulphur. 

Calcium  sulphate. 
Sulphides. 
Iron  pyrite. 

Seventh  Group. 

Calcium  fluoride. 
Sodium  chloride. 
Potassium  chloride. 
Sodium  bromide. 
Potassium  iodate. 
Manganese  dioxide. 

Eighth  Group. 

Ferrous  carbonate. 

Ferric  oxide. 

Nickelous  carbonate. 

Copper  carbonate. 

Copper. 

Silver. 

Silver  sulphide. 

Silver  chloride. 

Gold. 

Platinum. 


-176- 


Review  Exercise. 

163.  With  a  view  of  fixing  in  the  student's  mind  the  natural 
forms  in  which  the  common  elements  occur,  as  well  as  the 
processes  by  which  they  are  transformed  to  the  common  artifi- 
cial compounds,  the  following  exercise  has  been  designed: 

Table  II  gives  some  of  the  natural  sources  of  the  common 
elements,  and  Table  III  gives  the  names  of  many  common 
elements  and  compounds. 

The  student  is  assigned  some  element  or  compound  from 
Table  III,  by  number,  and  he  is  expected  to  start  with  natural 
elements  or  compounds  from  Table  I  and  state  in  words  what 
operations  are  necessary  to  produce  the  given  substance.  He 
is  not  allowed  to  make  use  of  any  substance  whose  formation 
from  the  natural  substances  he  has  not  first  shown.  He  can 
best  represent  all  changes  by  equations,  and  then  in  his  recita- 
tion translate  his  equations  into  words. 

The  value  of  this  exercise  cannot  be  overestimated.  It 
results  in  fixing  in  the  mind  the  relative  importance  of  certain 
artificial  compounds  and  fundamental  processes  in  a  way  that 
no  ordinary  laboratory  or  text-book  work  can  do. 

Example.  Show  the  artificial  preparation  of  cupric  car- 
bonate. 

Natural  materials.  Copper ;  Sulphur ;  Oxygen ;  Water ; 
Calcium  carbonate;  Sodium  chloride. 

S  +  02          =  S02. 

502  +  O      =  SO3. 

503  +  H2O  =  H,SO4. 

2H2SO4  +  Cu  =  CuSO4  +  SO2  +  2H2O. 
H2SO4  +  2NaCl  =  Na2SO4  +  2HC1.  " 
Na2SO4  +  2C      =  Na2S  +  2CCX. 
Na2S  +  CaCO3    =  Na2CO3  +  CaS. 

Na2CO3  +  CuSO4  =  CuCO,  +  Na2SO4. 

Here  again  the  student  is  not  expected  to  invent  any  reac- 
tions, but  to  use  his  text  book  freely  to  find  out  what  actual 
processes  are  in  use. 


—177— 
TABLE    III. 

Practice  Exercise. 

164.  Indicate  by  equations  the  formation  of  each  of  the 
following  compounds  or  elements  from  the  natural  elements  or 
compounds,  as  given  in  a  previous  table.  Use  no  reagents 
whose  formation  from  natural  substances  you  have  not  shown. 


1.  Sulphuric  acid. 

2.  Hydrochloric  acid. 

3.  Nitric  acid. 

4.  Calcium  oxide. 

5.  Carbon  dioxide. 

6.  Calcium  hydroxide. 

7.  Sodium  hydroxide. 

8.  Sodium  carbonate. 

9.  Sodium  bicarbonate. 

10.  Ammonia. 

11.  Hydrogen  sulphide. 

12.  Cupric  oxide. 

13.  Cuprous  oxide. 

14.  Cupric  sulphate. 

15.  Ferrous  chloride. 

1 6.  Ferric  hydroxide. 

17.  Ferrous  sulphate. 

1 8.  Hydrogen. 

19.  Iodine. 

20.  Bromine. 

21.  Aluminum 

22.  Ammonium 

23.  Cadmium 

24.  Calcium 

25.  Cupric 

26.  Cuprous 

27.  Magnesium 

28.  Nickelous 

29.  Potassium 

30.  Sodium 

31.  Lead 

32.  Barium 

33.  Zinc 

34.  Acetic  acid. 

35.  Arsenic  acid 

36.  Boric  acid. 

37.  Hydrobromic  acid. 

38.  Hydriodic  acid. 

39.  Hydrofluoric  acid. 

40.  Phosphoric  acid. 


bromide. 

61. 

acetate. 

62. 

chlorate. 

63- 

hydroxide. 

64. 

nitrate. 

65- 

carbonate. 

66. 

sulphide. 

67. 

sulphate. 

68. 

phosphate. 

69. 

oxide. 

70. 

nitrite. 

7i- 

sulphite. 

72. 

chloride. 

73- 

74- 

75- 

76. 

:id. 

77- 

78. 

id. 

79- 

L 

80. 

41.  Antimony. 

42.  Barium. 

43.  Copper. 

44.  Tin. 

45.  Lead. 

46.  Manganese. 

47.  Mercury. 
Silver. 
Iron. 
Zinc. 

Aluminum. 
Ammonium  chloride. 

53.  Ammonium  arsenate. 

54.  Antimony  sulphide. 

55.  Barium  chromate. 

56.  Chromic  sulphate. 

57.  Hydrogen. 

58.  Hydrogen  peroxide. 

59.  lodic  acid. 

60.  Lead  chromate. 

61.  Manganous  sulphate. 
Mercuric  chloride. 
Mercurous  chloride. 
Mercuric  oxide. 
Mercurous  oxide. 
Mercuric  nitrate. 
Mercurous  nitrate. 
Phosphorus. 
Potassium  arsenate. 
Potassium  dichromate. 
Potassium  bicarbonate. 
Potassium  chromate. 
Potassium  permanganate. 
Silver  nitrate. 

Silver  chloride. 
Sodium  arsenate. 
Sodium  thiosulphate. 
Stannous  sulphide. 
Stannous  chloride. 
Stannic  chloride. 


48. 
49- 
50- 
5i. 

52 


-178- 

TABLE   IV. 

165.  Showing  the  solubility  of  some  compounds  in  water. 
The  numerator  shows  the  number  of  parts  soluble  in  100  parts 
of  water  at  100°  C,  the  denominator  the  number  of  parts  soluble 
in  loo  parts  of  water  at  20°  C. 


Bromide. 

Carbonate. 

Chloride. 

Chromate. 

u 

Ferrocyanide 

Ferricyanide.  • 

Hydroxide. 

Iodide. 

Nitrate. 

(i 
a 

o 

Phosphate. 

"5, 

1 

1 

Al 

s 

S 

I 

's 

•I 

I 

89.1* 

36.1 

NH4  -- 

S 

S 

77-3 
37-2 

S 

S 

S 

S 

s 

S 

Dec. 

185 

... 

s 

103-3 

75-4 

S 

Ba  .... 

149 
104 

I 

72 
42 

I 

s 

S 

S 

5 

S 

~876 

90+ 
3-4 

I 

Dec. 
Dec. 

Ca  .... 

S 

I 

S 

S 
•  4 

Dec. 

s 

S 

.066 
•  15 

435+ 
204 

S 

^Dec. 

•V,.   .< 

Is 

.21 

.24' 

.27 
•23 

Cu'  ... 

I 

... 

I 

... 

I 

I 

I 

I 

I 

... 

I 

... 

... 

... 

Cu" 

s( 

S 

I 

I 

I 

I 

Dec. 

S 

j 

I 

203 

I 

42  •, 

Fe"... 

s 

I 

s 

I 

I 

I 

S 

S 

I 

I 

4_2 

I 

S 

24 

Fe"'.._ 

s 

... 

S 

... 

... 

I 

S 

I 

... 

s 

I 

I 

S 

I 

Pb_... 

s 
s 

I 

s± 

•97 

I 

s 
s 

I 

s 
s 

I 

•  14 
.04 

7 

52 

I 

I 

I 

I 

Mg  ... 

s 

I 

S 

S 

... 

s 

S 

I 

S 

s 

I 

I 

73 
36 

Dec. 

Hg'  ... 

1 

... 

I 

s 
s 

... 

... 

... 

I 

I 

Dec. 

I 

Dec. 
I 

.2 

I 

Hg"- 

4 
•4 

... 

54 
7 

Dec. 

40 

IO 

... 

... 

I 

I 

S 

I 

s 
I 

Dec. 

I 

102 

147+ 

56. 

79 

I00± 

77 

209 

247 

c 

26 

64 

112 

35- 

62 

35± 

5°± 

144 

31 

O 

ii 

Ag—  . 

I 

I 

I 

I 

I 

I 

I 

I 

I 

227 

I 

I 

i 
•5 

I 

Na—  . 

120 

"88 

45* 

21 

40 
36 

S 

s 

~&± 

66 

20 

S 

312 

178 

i    * 

Dec. 

s 

4I2t 

140 

S 

c_ 

200± 

I1 

102* 

Dec. 

18 

370 

101 

2 

or  

99 

, 

54 

S 

.68 

179 

71 

.6 

Sn"  ... 

S 

I 

S 

... 

... 

I 

... 

I 

s 
s 

— 

I 

I 

Dec. 

I 

Zn  .... 

s 

I 

S 

... 

I 

I 

... 

I 

S 

S 

I 

I 

653 
161 

I 

*  Anhydrous. 
S  Very  soluble, 
s  Slightly  soluble. 


t  At  34°  C. 

I  Insoluble. 
Dec.  Decomposed  by  boiling. 


—179— 

Solubility  Rules. 

These  rules  should  be  memorized. 

1 66.  All  the  hydrogen,  sodium,  potassium  and  ammonium 
compounds  are  soluble;  also  all  the  acetates,  chlorates  and 
nitrates. 

All  the  chlorides  are  soluble  except  those  of  silver,  lead  and 
mercurous  mercury. 

All  the  sulphates  are  soluble  except  those  of  strontium, 
barium  and  lead. 

All  the  carbonates  and  phosphates  are  insoluble  except  those 
of  sodium,  potassium  and  ammonium. 

All  the  hydroxides  are  insoluble  except  those  of  sodium, 
potassium,  ammonium,  calcium,  strontium  and  barium. 


—ISO- 
TABLE   V. 
PERIODIC   ARRANGEMENT   OF   THE   ELEMENTS. 

167.     A  few  of  the  very  rare  elements  are  omitted. 
H  =  1.0075. 


0 

I 

II 

III 

IV 

V 

VI 

VII 

VIII 

R 

RH 
R2O 

RH2 
RO 

RH3 
R203 

RH4 
RO2 

RH3 
R305 

RHa 
R03 

RH 
R307 

RO4 

He 
4 

Li 

7.03 

Gl 
9.1 

B 

II.  0 

C 

12.0 

N 
14.04 

0 

16.00 

F 
19 

Ne 

20 

Na 
23-05 

Mg 
24.36 

Al 
27.1 

Si 

28.4 

P 

31.0 

S 
32.06 

Cl 

35-45 

A 

39-9 

K 
39-15 

Ca 
40.1 

Sc 
44-1 

Ti 

48.1 

V 

51.2 

Cr 
52.1 

Mn 

55-0 

Fe        Co        Ni 
55-9      59-0       58.7 

Cu 
63.6 

Zn 

65.4 

Ga 
70 

Ge 

72.5 

As 

75-0 

Se 
79-2 

Br 
79-90 

Kr 

81.8 

Rb 

85.5 

Sr 
87.6 

Yt 

89 

Zr 

90.6 

Cb 
94 

Mo 

96 

Ru        Rh       Pd 
101.7    I03        106.5 

Ag 
108.0 

Cd 
112.4 

In 
H5 

Sn 
119.0 

Sb 

120.2 

Te 
127.6 

I 
127.0 

Xe 

128 

Cs 
132.9 

Ba 
137-4 

La 

138.9 

Ce 
140.25 

Ta 

183 

W 

184 

Os         Ir         Pt 

191         193       194.8 

Au 
197.2 

Hg 

200 

Tl 
204.1 

Pb 
206.9 

Bi 

208.5 

Ra 

225 

Th 
232-5 

u 

238.5 

-i83- 


i6g.  TABLE  VII. 

TABLE  OF  HEATS  OF  FORMATION  OF  SOME  COM- 
MON COMPOUNDS  AT  NORMAL  TEMPERATURE. 


Heat  of  I 

"ormation. 

Substance. 

Gaseous. 

Liquid. 

Solid. 

Dissolved., 

Ammonia.         _                  .  . 

+  12.  0 

+    2O  4 

Ammonia  chloride  

+    75.8  , 

+    71.  Q 

Ammonium  nitrate 

+   88  o 

4-    8l   8 

Barium  carbonate- 

+  280  5 

Barium  chloride 

+  IQ4  7 

+  196  8 

Barium  oxide                     -   - 

+  124  2 

-f  TC8    7 

Calcium  carbonate  .. 

+  267  7 

Calcium  hydroxide 

+  214  2 

Calcium  oxide 

+  131  o 

+  I4O  ^ 

Carbon  dioxide             -   . 

+  Q7  6 

+  101  5 

Carbon  disulphide  

—  28.7 

—    22  3 

Cupric  chloride 

+    5l6 

+   62  7 

Cupric  oxide                  -   - 

+    -57  2 

Hydrochloric  acid     _   

+  22.  0 

+      -JQ    -7 

Hydrogen  sulphide  _         

+    2.7 

-f      7-7 

Lead  oxide  (PbO) 

+      CQ    "3, 

Lead  sulphide 

+    184 

Magnesium  chloride           

+  1510 

+  186  q 

Magnesium  oxide  _• 

• 

+  144.0 

Magnesium  sulphate 

+  3O2  1 

+  ^2Q  O 

Nitric  acid 

+    41  Q 

+    4Q   I 

Nitrogen  pentoxide 

-f-       TO     T 

+    2O  8 

Potassium  carbonate 

+  278  4 

+  284  Q 

Potassium  chloride 

+  105  6 

+  IOI  2 

Potassium  hydroxide  

+  103.2 

+  116  5 

Potassium  nitrate 

+  no  ^ 

+  III  O 

Potassium  oxide 

+  164  6 

Potassium  sulphate            

+  344  6 

+  q^8  2 

Sodium  bicarbonate 

+  227  o 

4-  22^  7 

Sodium  carbonate 

+  26Q  Q 

+  275  4 

Sodium  chloride                  

-f    07  6 

4-     QO   J. 

Sodium  hydroxide 

4-  IOI  Q 

+  in  8 

Sodium  nitrate 

+  III3 

-f  1  06  3 

Sodium  oxide 

+  IOO  2 

+  ice  2 

Sodium  sulphate 

+  ^28  4 

+  320  o 

Sulphur  trioxide 

+  IO^  ^ 

+  142  ^ 

Sulphuric  acid 

+  180  Q 

+  210  9 

Water     

+  58.O 

+  68.  o 

4-    60.8 

Zinc  sulphate 

4-  230  o 

+  248  o 

— 184— 


170.  TABLE   VIII. 

Tables  of  Length,  Weight  and  Volume. 

10  millimeters  =  one  centimeter. 
100  centimeters  =  one  meter. 
I  meter  ==  39.37  inches. 

I  cubic  centimeter  of  pure  water  at  a  temperature  of  4°  C. 
weighs  one  gram. 

looo  grams  =  i  kilogram. 
I  kilogram  =  2.2  pounds. 
looo  cubic  centimeters  =r  i  liter. 
I  liter  =  i. 06  quarts. 

Exact  Equivalents. 

i  inch  =  25.39954  millimeters. 

I  liter  —  35.197  fluid  ounces. 

i  ounce  Avoirdupois  =  28.34954  grams. 

i  gram  =  15.43235  grains. 


-i85- 

LABORATORY  EQUIPMENT. 
Student's  Individual  Equipment  with  approximate  cost. 

171.     i  Beaker,  No.  2  Griffin's,  with  lip,  Bohemian  glass  $  .10 

1  Long  right  angle  bend,  8  cm.  X  45  cm 02 

3  Medium  right  angle  bends,  8  cm.  X  15  cm 06 

3  Short  right  angle  bends,  8  cm.  X  8  cm 04 

2  500  c.c.  bottles,  3.5  cm.  across  the  mouth 10 

6  5  oz.  flint  glass  quinine  bottles,  3.5  cm.  across  the 

mouth.  Whitall  Tatum  Co 18 

i  Combustion  spoon,  y2  inch 15 

i  Royal  Berlin  porcelain  evaporating  dish,  3  inch..  .25 

i  File,  three  cornered,  4  inch 10 

I  Carbon  dioxide  flask,  holding  285  c.c.  when  filled 

to  the  brim,  and  measuring  3.5  cm.  across  the 

mouth  15 

I  Pair  of  iron  forceps,  4.5  inch.  Eimer  &  Amend 

new  catalogue,  No.  3335 09 

i  Funnel  of  glass,  2.^/2  inch 05 

i  Funnel  tube,  18  inch 10 

8  Squares  of  glass,  two  inches  square 08 

i  Iron  wire  loop.  No.  18  standard  wire  gauge.  . .  .01 
i  Mortar  and  pestle,  E.  &  A.,  No.  4093,  No.  7  23/2 

inch  25 

i  Glass  stirring  rod 01 

}4  lb.  soft  German  glass  tubing,  inside  diam.,  $l/2 

mm.,  outside  diam.,  7^  mm 10 

3  2*/2  inch  rubber  connectors,  made  from  *4  inch 

rubber  tubing,  double  thick,  white  stock 06 

i  8  inch  piece  of  3/16  inch  black  rubber  tubing  of 

pure  gum,  double  thick 08 

3  No.  8  two-hole  rubber  stoppers,  small  diam.,  33 

mm 45 

i  No.  3  one-hole  rubber  stopper 05 

12  Test  tubes,  6  inch  X  24  inch 24 

i  Wire  test  tube  holder 10 

i  Wooden  test  tube  rack 25 

3  Pieces  of  copper  wire 01 


— 186— 

Equipment  for  general  use. 

One  set  of  the  following  material  should  be  constantly  avail- 
able at  every  laboratory  position : 
172.     1000  c.c.  wide  mouth  bottle $  .10 

1  bulb    test   tube.      Eimer   &   Amend,    No.    4877. 

These  tubes  are  a  little  too  thick  on  the  bottom 
and  are  much  improved  by  having  a  second 
small  bulb  blown  on  the  end 10 

2  Bunsen  burners  with  tubing 70 

i  Foot  rule  with  centimeter  scale 10 

i   Square  of  cobalt  glass,  4"  X  4" 10 

i  500  c.c.  filter  flask  with  tubing 20 

i  Package  of  cut  filter  paper,  10  cm.  diam 10 

i  Brass  filter  pump,  E.  &  A.,  3250 1.25 

Filter  paper   discs.      These  may  best  be   cut 

from   ordinary   sheet   filter   paper   with   a   steel 

punch. 
I  50  c.c.  graduated  cylinder,  double  graduation.  . .       .50 

I  2  inch  iron  crucible,  E.  &  A.,  2879 3° 

i  2  inch  square  iron  plate  with  depression  in  center       .05 
i  Lamp  stand  of  iron.     Rod  18  inches  high  and  y% 

inch  diam 35 

i  3  inch  ring  for  stand .05 

i  2  inch  ring  for  stand 05 

i  Clamp  for  stand,  E.  &  A.,  2750 40 

i  Package  of  wrapping  paper 05 

i  2  inch  porcelain  sieve  funnel,  E.  &  A.,  No.  3357       .50 

i  Rubber  stopper  for  funnel 10 

i  Litmus  paper  bottle 15 

i  8  oz.  Cone,  sulphuric  acid  bottle 23 

i     "       Dil.  "  "         "     23 

i     "       Cone,  nitric  acid  bottle 23 

I     "       Dil.          "         "         "     23 

I     "       Cone,  hydrochloric  acid  bottle 23 

i     "       Dil.  "  "         "     23 

i     "       Ammonium  hydroxide  bottle 23 

i     "       Sodium  hydroxide  bottle 23 

These  bottles  have  glazed  white   labels   with 

transparent  letters  and  formulas,  Whitall  Tatum 

Company. 


-i87- 

I  Tin  cup  for  shelf  for  water  pan,  with  hole  in 

bottom  and  side $  .05 

i  Common  centigrade  thermometer,  E.  &  A.,  4894 

graduated  up  to  130°  C 50 

I  Bundle  of  tooth  picks  for  splints 

I  Water  pan,  any  kind  of  a  rectangular  pan  over 

4  inches  deep 15 

i  Iron  wire  gauze  with  asbestos  center,  E.  &  A., 

7262     12 

i  Voltameter,  see  special  fig.  4,  made  to  order  by 

Bausch  &  Lomb,  duty  free,  glass  part  only. .  .  2.00 

i  Harvard  trip  scale,  E.  &  A.,  2142 7.00 

i  Set  of  weights,  E.  &  A.,  2204,  i  g.  to  500  g.  .  . .  2.25 

1  Test  tube  cleaner  with  brush  on  end. .  . .10 

Additional  Apparatus  of  Use  in  Making  up  Solutions  and 

in  Making  Preparations. 
Hot  plate   $5.00 

2  8  liter  flasks i  .50 

2  2.l/2.  liter  flasks 80 

i   12  inch  Royal  Berlin  porcelain  evaporating  dish     2.00 

i  5  inch  glass  funnel 30 

i  Royal  Berlin  porcelain  casserole,  No.  6 1.50 

i  6  inch  porcelain  sieve  funnel,  E.  &  A.,  3356.  . .  .     2.00 
6  No.  6  Griffin's  beakers  with  lip 1.20 

Folded  filter  paper,  38  cm 

i   Sodium  press 12.00 

Reagents  sufficient  to  enable  a  class  of  twenty  students  to 
perform  the  experiments  in  this  book,  with  approximate 
cost. 

173.  Alcohol,  i  pint $  .35 

Potassium  alum,  l/2  Ib 10 

Aluminum  sulphate,  i  Ib.,  c.p 25 

Aluminum  wire,  I  oz 10 

Aluminum  foil,  i  oz 10 

Ammonium  chloride,  y2  Ib 12 

Ammonium  nitrate,  ^  Ib 25 

Ammonium  thiocyanate,  i  oz 05 

Ammonium  sulphate,  com.,  2  Ib 20 


— 188— 

Ammonium  sulphide,  made  by  passing  hydrogen 
sulphide  into  ammonium  hydroxide  until  sat- 
urated and  then  adding  an  equal  volume  of 
ammonium  hydroxide. 

Arsenic,  metal,  j£  Ib $  .30 

Arsenic  trioxide,  >4  Ib 10 

Borax,  i  Ib 10 

Bromine  water,  made  by  shaking  liquid  bromine 
with  water  and  keeping  an  excess  of  bromine 
in  bottle. 

Bromine,  2  oz 25 

Barium  chloride,  c.p.,  2  oz 05 

Barium  nitrate,  com.,  ^  Ib 10 

Cadmium  sulphate,  I  oz 20 

Calcium  chloride,  com.,  y2  Ib 05 

Calcium  nitrate,  com.,  i  oz 15 

Calcium  oxide,  lime,  2  Ib 05 

Carbon  disulphide,  2  Ib 44 

Charcoal,  animal,  i  Ib 02 

Charcoal,  wood,   i   Ib 01 

Chloride  of  lime,  i  Ib 05 

Chlorine  water 

Chrome  alum,  %  Ib 05 

Coal,  bituminous,  3  oz 

Copper,  sheet,  i  Ib 40 

Copper  wire,  No.  18  standard  wire  gauge,  %  Ib.  . .        10 

Copper  gauze,  80  mesh,  10  grams 20 

Copper  scale,  y^  Ib .10 

Copper  sulphate,  com.,  i  Ib 10 

Cream  of  tartar,  i  Ib 35 

Ferrous  sulphate,  com.,  2  Ib 06 

Ferrous  sulphide,  com.,  lumps,  5  Ib 50 

Grape  sugar,  i  oz 02 

Hydrofluoric  acid,  4  oz 50 

Iron  filings,  ^2  Ib 05 

Iron  wire,  fine,  2  Ib 10 

Iron  wire,  2  Ib.  No.  16  standard  wire  gauge 20 

Iodine,  i  oz 25 

Lead  wire,  i  oz 05 

Lead  nitrate,  com.,  i  oz .02 


-i89- 

Lead  oxide,  litharge,  J4  ^ $  .03 

Lead  oxide,  red  lead,  i  oz 03 

Magnesium  wire,  I  oz 55 

Magnesium  oxide,  ^  lb 25 

Manganese  dioxide,  I  lb.,  powder 08 

Marble,  I  lb. 10 

Mercury,  I  oz 10 

Mercuric  chloride,  2  oz 20 

Mercuric  oxide,  red,  2  oz 20 

Oxygen  mixture,  made  by  mixing  two  parts  of  pul- 
verized potassium  chlorate  with  one  part  of 

powdered  manganese  dioxide,  I  lb 30 

Phosphorus,  red,  I  oz 10 

Phosphorus,  yellow,  I  oz 10 

This    phosphorus    should    be    granulated    as 
described  under  Laboratory  Suggestions.     176. 

Potassium  bromide,  c.p.,  2  oz 20 

chloride,  c.p.,  ^  lb 15 

chlorate,  c.p.,  y2  lb 10 

chromate,  I  oz 05 

dichromate,  y2  lb 07 

ferrocyanide,  c.p.,  I  oz 02 

ferricyanide,  c.p.,  I  oz 05 

hydroxide,   pure   by   alcohol   in   sticks, 

I    OZ IO 

iodide,  c.p.,  I  oz 30 

nitrate,  com.,  I  lb 05 

permanganate,  I  oz 05 

sulphate,  I  oz 05 

tartrate,  normal  salt,  */£  lb 40 

Quinine  sulphate,  %  oz 10 

Sand,  i  oz 

Silver  nitrate,  I  oz 5° 

Silver  wire  or  foil,  6  g 20 

Sodium  bicarbonate,  c.p.,  i  lb 20 

carbonate,  soda  ash,  2  Ibs 10 

chloride,  table  salt,  2  Ibs 05 

nitrate,  com.,  i  lb 10 

nitrite,  com.,  i  lb 10 


— IQO 

Sodium  phosphate,  common  cryst,  I  Ib $  .10 

sulphate,  com.  Glauber  salt,  I  Ib 03 

thiosulphate,  I  Ib 05 

Sodium,  metal,  i  oz 25 

Sodium  wire, — this  is  made  when  needed  by  means 
of  a  sodium  press. 

Starch,  corn  starch,  I  Ib 10 

Stannous  chloride,  made  when  needed  by  dissolving 
tin  in  hydrochloric  acid. 

Strontium  nitrate,  com.,  l/2  Ib 07 

Sulphur,  brimstone,  3  Ibs 15 

Tartaric  acid,  y2  Ib 15 

Tin,  granulated,  y^  Ib 15 

Water  glass,  ^  Ib 05 

Zinc,  granulated,  2  Ib 40 

oxide,  l/4  Ib 03 

sulphate,  ^4  Ib 02 

Stock  Acids  and  Solutions  for  a  class  of  twenty,  for  which  it 
is  desirable  to  have  special  reagent  bottles. 

174.  Sulphuric  acid,  cone.,  c.p.,  specific  gravity  1.84, 

cont.  98%  acid,  10  Ibs $  .80 

Sulphuric  acid,  dil.,  made  by  mixing  one  part  of  the 
cone,  acid  with  three  parts  of  distilled  water, 
s.g.  1.28,  cont.  38.13%  acid. 

Nitric  acid,  cone.,  c.p.,  s.g.  1.42,  8  Ibs 80 

Nitric  acid,  dil.,  made  by  mixing  the  cone,  acid 
with  an  equal  volume  of  water,  s.g.  1.28,  cont. 
45.55%  acid. 

Hydrochloric  acid,  cone.,  c.p.,  s.g.  1.2,  cont.  40% 

acid,  8  Ibs 80 

Hydrochloric  acid,  dil.,  made  by  mixing  equal  vol- 
umes of  the  cone,  acid  and  distilled  water,  s.g. 
i.i,  cont.  19.8%  acid. 

Ammonium  hydroxide,  cone.,  c.p.,  s.g.  .90,  2  Ibs.  .  .       .20 

Ammonium  hydroxide,  dil.,  made  by  mixing  one 
volume  of  cone,  ammonia  with  three  vols.  of 
water,  s.g.  .97  and  cont.  15%  ammonium 
hydroxide. 


Sodium  hydroxide  solution,  cone.,  made  by  dissolv- 
ing 1000  g.  commercial  granulated  sodium 
hydroxide  in  1000  c.c.  of  water  in  an  iron  dish, 
allowing  the  solution  to  settle  for  a  week  and 
siphoning  off  the  clear  solution;  s.g.  1.51 
and  cont.  51%  sodium  hydroxide.  2  Ibs.  crude 
NaOH  $  .20 

Sodium  hydroxide  solution,  dil.,  made  by  diluting 
one  volume  of  the  cone,  solution  with  three 
vols.  of  water;  s.g.  1.18  and  cont.  16% 
NaOH. 

Alcohol,  ethyl,  95%. 

Ammonium  sulphide. 

Barium  chloride,  10%  solution. 

Bromine  water,  saturated  with  bromine. 

Calcium     hydroxide,     saturated     solution     made     when 
wanted. 

Carbon  disulphide. 

Chlorine  water.     Made  when  wanted. 

Copper  sulphate,  10%  solution. 

Lead  nitrate,  10%  solution. 

Litmus  solution,  made  by  dissolving  about  I  g.  of  azolit- 
min  in  two  liters  of  distilled  water  and  filtering. 

Mercuric  chloride,  5%  solution. 

Potassium  chromate,  10%  solution. 

Potassium  dichromate,  10%  solution. 

Potassium  ferrocyanide,  10%  solution. 

Potassium  iodide,  $%  solution. 

Potassium  permanganate,  10%  solution. 

Silver  nitrate,  5%  solution. 

Sodium  phosphate,  10%  solution. 

Stannous  chloride,  10  g.  of  tin  to  100  c.c.  of  acid. 


— 192- 

Suggestions  for  the  Laboratory. 

175.  Hoods.  There  should  be  one  hood  for  every  four 
students  in  a  laboratory  class. 

Lockers.  Each  student  should  have  an  individual  locker 
with  a  combination  lock. 

Distribution  of  Gases.  If  possible  there  should  be  an  inde- 
pendent pipe  system  with  an  outlet  at  each  position  so  arranged 
that  any  gas  such  as  oxygen,  hydrogen,  carbon  dioxide  or  air 
could  be  delivered  at  a  moment's  notice. 

Ordinary  ^4 -inch  gas  pipe  is  large  enough  for  the  purpose. 

Oxygen.  This  should  be  made  as  wanted  and  kept  in  a  large 
gas  holder. 

Hydrogen.  Should  be  furnished  by  a  large  Parsons'  gen- 
erator. E.  &  A.  3611.  This  generator  has  been  used  in  this 
laboratory  for  six  years,  and  has  given  perfect  satisfaction. 
It  has  run  for  one  year  without  any  attention  whatever,  the 
gas  being  used  intermittently. 

Carbon  Dioxide.  This  gas  might  well  be  made  in  a  Parsons' 
generator,  but  it  is  much  cheaper  to  buy  it  in  liquid  form  in 
large  cylinders  and  distribute  it  through  a  reducing  valve. 

Hydrogen  Sulphide.  A  Parsons'  generator  should  be  set 
up  in  the  basement  or  attic  and  connected  with  each  hood  by 
means  of  lead  or  iron  pipe.  It  requires  little  or  no  attention 
during  the  year.  At  the  most  a  mere  refilling  of  the  reservoir. 

Sulphur  Dioxide.  This  should  be  bought  in  liquid  form  in 
tin  cylinders  holding  about  five  pounds.  But  to  avoid  accidents 
due  to  defective  cylinders  the  whole  should  be  inclosed  in  a 
piece  of  iron  pipe  with  a  screw  cap  on  each  end,  through  one 
of  which  the  brass  cock  of  the  cylinder  projects. 

Chemicals.  These  should  be  pure  as  far  as  possible.  It 
does  not  pay  to  use  impure  acids.  Buy  all  chemical  supplies 
direct  from  the  manufacturer.  Baker  &  Adamson,  of  Easton, 
Pa.,  furnish  very  satisfactory  chemicals  at  very  reasonable 
prices. 

176.  Granulated  Yellow  Phosphorus.  This  cannot  be 
bought,  but  may  be  readily  made  in  the  laboratory  as  follows : 

Stand  a  nine  liter  flask  having  a  cylindrical  neck  at  least  two 
inches  in  diameter  in  the  sink,  and  fill  it  with  cold  water  up  to 
the  beginning  of  the  neck. 


—193— 

Contract  the  end  of  a  one-inch  carbon  filter  tube  by  heating, 
to  a  one-eighth  inch  hole.  Push  this  end  barely  through  the 
center  of  a  disc  of  sheet  cork  just  a  little  larger  than  the  diameter 
of  the  neck  of  the  flask.  Now  push  the  cork  in  a  horizontal 
position  down  into  the  neck  of  the  flask  until  the  filter  tube  is 
entirely  within  the  neck,  the  top  of  the  tube  being  about  an 
inch  below  the  top  of  the  flask.  Add  cold  water  through  the 
tube  until  the  surface  reaches  the  cork,  and  fill  the  neck  with 
boiling  water.  Cut  up  all  the  stick  phosphorus  into  pieces  not 
more  than  one  inch  long,  and  quickly  drop  them  one  at  a  time 
into  the  filter  tube.  They  will  melt  almost  at  once  and  run  down 
into  the  flask  in  the  form  of  shot.  The  first  to  go  through  will 
probably  go  into  the  undercooled  condition  and  operations  must 
be  suspended  until  they  crystallize.  Add  boiling  water  to  the 
neck  of  the  flask  from  time  to  time.  A  pound  of  phosphorus  can 
be  made  into  thousands  of  pellets  in  this  way  in  less  than  half 
an  hour.  Reasonable  caution  must  be  used  in  handling  so 
much  phosphorus. 

Bunsen  Burner  Igniter.  A  single  wire  circuit  should  be 
run  to  a  pointed  brass  contact  piece  on  the  shelf  in  front  of 
every  Bunsen  burner  in  the  laboratory.  A  ten-cell  dry  battery 
and  spark  coil  are  put  in  the  circuit,  and  the  other  end  con- 
nected to  the  gas  pipe  system.  A  loose  bare  copper  wire  is 
run  through  every  rubber  tube  and  projects  into  the  gas  nipple 
and  burner  nipple. 

It  is  only  necessary  to  turn  on  the  gas  and  touch  the  top  of 
the  burner  to  the  contact  piece  to  get  a  light. 

This  scheme  has  been  in  use  in  this  laboratory  for  about 
four  years  and  is  perfectly  satisfactory. 

It  would  be  a  still  better  plan  to  run  a  two-wire  system  and 
use  a  double  contact  piece,  the  circuit  being  closed  by  the  tip 
of  the  burner.  If  this  were  done,  the  no  volt  current  could 
be  turned  on  through  two  32  candle  power  lamps  abreast  and 
a  much  hotter  spark  would  be  obtained. 

Suction  Pump.  Every  position  should  be  provided  with  a 
suction  pump  and  a  filter  flask  with  a  small  porcelain  sieve 
funnel. 

Master  Key  for  Lockers.  Have  each  combination  lock 
engage  with  a  common  lock  set  in  the  jamb.  All  the  common 


—I94— 

» 
locks  being  alike,  one  key  in  the  hands  of  the  instructor  will 

enable  him  to  open  any  locker  in  the  laboratory  instantly. 

Glass  Tubing.  This  should  be  of  the  soft  German  variety, 
and  the  standard  size  for  all  glass  tubes  used  in  the  experiments 
should  be  7^2  mm.  outside  diam.  and  5  mm.  inside  diam. 

Dry  Reagents.  All  dry  reagents  that  are  regularly  issued 
to  the  students  and  are  not  altered  by  exposure  to  the  air 
should  be  kept  in  5-oz.  wide-mouth  bottles,  as  many  of  each 
kind  as  there  are  positions.  And  when  issued  there  should  be 
a  small  horn  spoon  in  each  bottle. 

Horn  Spoons.  These  are  double  horn  spoons  5  inches  long, 
E.  &  A.  No.  4706.  They  cost  about  six  cents  apiece,  duty  free. 

Flasks  and  Bottles.  All  flasks  and  bottles  should  be  selected 
with  the  same  sized  openings  in  order  that  only  one  size  of 
rubber  stopper  need  be  used. 

Care  of  Wooden  Tables. 

Solution  A.  Shake  up  Bismark  brown  with  wood  alcohol 
in  a  gallon  bottle.  100  grams  of  the  color  to  a  quart  of  alcohol. 

Solution  B.  A  solution  of  orange  shellac,  best  obtained  at 
the  painter's. 

Solution  C.     Made  by  mixing  equal  volumes  of  A  and  B. 

Solution  D.  Dissolve  one  or  two  ounces  of  beeswax  in  one 
quart  of  boiled  linseed  oil  with  the  aid  of  heat  in  an  iron  dish 
or  pot.  Make  this  up  to  a  gallon  with  turpentine. 


Paint  the  tables  all  over  with  solution  C  and  from  time  to 
time  in  spots  where  they  need  it.  About  once  a  month  rub 
them  all  over  with  the  oil  wax  polish,  solution  D. 

The  Bismark  brown  is  quickly  turned  to  a  light  color  by 
alkalies  but  the  color  may  be  quickly  restored  by  rubbing  the 
spot  with  dilute  nitric  acid,  half  water  and  half  strong  acid. 
It  is  a  good  plan  to  rub  them  all  over  once  in  a  while  with 
the  dilute  nitric  acid.  This  changes  the  oil  to  a  very  hard 
hornlike  substance  and  makes  them  look  as  though  they  had 
just  been  revarnished. 


INDEX. 


PAGE. 

Acids    102 

Acids,   non-volatile    106 

unstable    106 

volatile  and  stable  106 

Acid  Salts,  Ex.  25    49 

Alkali     102 

Aluminum,  Ex.  33  61 

Ammonia,  Preparation  and  Properties,  Ex.  19   39 

Amorphous    18 

Analysis  89 

Anhydride,  Acid  103 

Apparatus,  Additional    187 

Arsenic,  Ex.  22   45 

Atom    86 

Atomic  Hypothesis  of  Daiton  91 

Atomic  Weight  87 

Atomic  Weight,   Derivation   of    123 

Avogadro's  Hypothesis    122 

Baking  Powder,  Ex.  25  49 

Barium,  Ex.  29  55 

Base 102 

Basic  Salt  49 

Binary  Compound   90 

Boiling  Point 132 

Boiling  Point  Law    133 

Boyle's   Law    129 

Bromine,  Preparation  and  Properties,  Ex.  15  31 

Bunsen  Burner,  Construction  and  Operation,  Ex.  I   9 

Bunsen  Burner  Igniter   192 

Cadmiun,  Ex.  30  56 

Calcium,  Ex.  29  55 

Calcium  Oxalate,  Solution  in  Hydrochloric  Acid   152 

Carbon,  Ex.  23    46 

Carbon  Dioxide   Generator    192 

Carbon  Dioxide,  Preparation  and  Properties,  Ex.  24  47 

Chemical  Change    86 

Composition    86 

Compounds,  Preparation  of,  Ex.  35 69 

Equilibrium   144 


196- 
PAGE 

Chemical  Equivalence  95 

Equation    96 

Equation,  Full  Meaning  of   98 

Equivalence  between  compounds  120 

Chlorine,  Preparation  and  Properties,  Ex.  13   28 

Chromium,  Ex.  33  61 

College  Examination  Questions   160 

Combining  Power   93 

Completed  Reactions 144 

Concentration   Cell    141 

Condensation    131 

Copper,  Ex.  31   57 

Critical  Pressure   132 

Critical  Temperature    132 

Crystalline    18 

Daniell  Cell  139 

Decant 22 

Diabasic  Acid    49 

Dissociation,  Degree  of,  in  solution   137 

Distillation    19 

Distribution  of  Gases  192 

Dry  Reagents   194 

Efflorescence    19 

Elements,  Alphabetical  List   161 

Electrodes    16 

Electrolysis  16,  139 

Electrolyte    16 

Element 85 

Epsom  Salt   24 

Equations  representing  formation  of  Binary  Compounds    97 

Equipment,   personal    i 

individual  Laboratory   I 

for  general  use   2,  186 

Evaporation    I31 

Exothermic  Reactions    142 

Faraday's  Law   I41 

Filtrate    n 

Flasks  and  Bottles   194 

Formula    88 

Formula,  derivation  of,  from  analysis,  and  vapor  density 89,  126 

Formula,  Full  Meaning  of  88 

Freezing  Point  Law    133 

Galvanic  Cell    J39 

Gas  Density    I23 

Gas  Density,  Determination  of  124 

Gaseous  Bodies,  Nature  of   128 

Gas  Volumes,  Reduction  of 130 


—197— 

PAGE 

Glass  Tubing    194 

Glass,  Working  with,  Ex.  2   9 

Glauber's  Salt  . , 23 

Heats  of  Formation,  Table   183 

Heat  of  Formation  and  Decomposition 142 

Heat  of  Reaction  143 

Hoods    192 

Horn  Spoons    194 

Hydrochloric  Acid,  Preparation  and  Properties,  Ex.  14 29 

Hydrogen  Generator  192 

Hydrogen  Sulphide   Supply    192 

Hydrogen  Sulphide,  Ex.  17  36 

Hydrogen,  Preparation  and  Properties,  Ex.  6 13 

Hydrolysis    153 

Iodine,  Preparation  and  Properties,  Ex.  16 31 

lonization    136 

Ions 136 

Iron,  Ex.  32   59 

Key  for  Lockers 193 

Laboratory  Equipment  185 

Law  of  Charles    130 

Law  of  Definite   Proportions, 90 

Law  of  Dulong  and  Petit   142 

Law  of  Multiple    Proportions    91 

Laws  of  Osmotic  Pressure 135 

Lead,  Ex.  27  52 

Liquefaction  132 

Lockers    192 

Magnesium,  Ex.  30  56 

Manganese,  Ex.  32   59 

Mass  Action  Law   144 

Matter   85 

Matter,  Properties  of  86 

Mechanical   Mixture    87 

Mercury,  Ex.  30   56 

Metallic  Oxides,  Reduction  of,  Ex.  8  18 

Metals    90 

Molecular  Weight    87 

Molecular  Weight,  Derivation  of   123 

Molecule    86 

Neutral    25 

Neutralization   25 

Nitric  Acid,  Ex.  20  41 

Nitrogen,  Oxides  of,  Ex.  20  41 

Nitrogen,  Preparation  and  Properties,  Ex.  19  39 

Non-metals    90 

Note-book  4 


— 198— 

PAGE 

Normal  Salt  49 

Normal  Temperatures 131 

Osmotic  Pressure 134 

Computation  of    135 

Outline  for  recitation   5 

Oxidation    109 

in  the  dry  way  109 

in  the  wet  way 114 

Oxide,  acidic  103 

alkaline    103 

basic    t  102 

Oxides  of  Elements,  in  groups in 

Oxidizing  agents    1 18 

Oxygen    192 

Oxygen,  Preparation  and  Properties,  Ex.  5  12 

Periodic  Arrangement  of  Elements 180 

Polybasic  Acid   49 

Potassium,  Ex.  28  54 

Phosphorus,  Ex.  22 45 

Physical  Change    86 

Precipitates  107 

formation  of   148 

prevention  of  the  Formation  of  151 

Precipitation  of  Metals  from  solution  139 

Precipitation  of  Soluble  Salts  from  Solution  149 

Problems    168 

Questions  on  Advanced  Theory  156 

on  Laboratory    Exercises 72 

on  Fundamentals    155 

Radical    99 

acid    99 

basic    99 

Radicals,  combinations  of 100 

Radicals,  names  of   99 

Reaction    96 

Reagent  96 

Reagents    187 

Reducing  Agents   119 

Reduction    119 

Residue   1 1 

Review  Exercise I72 

Review  Laboratory  Work,  Ex.  34   62 

Salt   102 

Salts,  Formation  of 104 

Salts,  Formation  of,  Ex.  10,  n,  12  23,  25,  27 

Silicon,  Ex.  26   51 

Silver,  Ex.  31 57 


—199— 

PAGE 

Sodium,  Ex.  28 54 

Solubility  Rules    179 

Solubility  Table    168 

Solution    21 

Solution  of  Metals  in  acids 138 

Specimen  Exercise    6 

Strontium,  Ex.  29 55 

Sublimation    32 

Suction  Pump  193 

Sulphur  Dioxide,    Liquid    192 

Sulphur,  Ex.  16  34 

Sulphur  Dioxide,  Ex.  17  36 

Sulphuric  Acid,  Preparation  and  Properties,  Ex.  18  37 

Symbol    85 

Symbol  of  an  Element,  Full  Meaning  of 88 

Tables  of  Length,  Weight  and  Volume    184 

Thermochemistry    142 

Tin,  Ex.  27   52 

Tribasic  Acid   49 

Valence    93 

Vapor  Density    123 

Water  of  Crystallization  18 

Water,  Purification  of,  Ex.  8 8 

Yellow  Phosphorus,  granulated    192 

Zinc,  Ex.  30   56 


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